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ADVANCED ALGEBRA 



BY 

HEEBEET E. HAWKES, Ph.D. 

Assistant Professor of Mathematics in Yale University 



X 



GINN & COMPANY 

BOSTON . NEW YORK • CHICAGO • LONDON 



QA\54 



e. 







Copyright, 1905, by 
H. E. HAWKES 



ALL BIGHTS KESERVED 

613.1 



GINN & COMPANY • PRO- 
PRIETORS . BOSTON . U.S.A. 



PEEFACE 

This book is designed for use in secondary schools and in 
short college courses. It aims to present in concise but clear 
form the portions of algebra that are required for entrance to 
the most exacting colleges and teclinical schools. 

The chapters on algebra to quadratics are intended for a 
review of the subject, and contain many points of view that 
should be presented to a student after he has taken a first 
course on those topics. Throughout the book the attention 
is concentrated on subjects that are most vital, pedagogically 
and practically, while topics that demand a knowledge of the 
calculus for their complete comprehension (as multiple roots, 
and Sturm's theorem) or are more closely related to other por- 
tions of mathematics (as theory of numbers, and series) have 
been omitted. 

The chapter on graphical r epresentation, has been intro- 
duced early, in the belief that the illumination which it affords 
greatly enlivens the entire presentation of algebra. The dis- 
cussion of the relation between , pairs of linear ftguntioT^ ^ g and 
pairs of straight lines is particularly su g gestive. _ 

In each chapter the discussion is directed toward a definite 

result. The chapter on theory of equations aims to give a 

simple and clear treatment of the method of obtaining the 

real roots of an equation and the theorems that lead to that 

iii 



383513 



vi CONTENTS 

CHAPTER n 
FACTORING 

SBOnOK ^ PAGE 

28. Statement of the Problem 16 

29. Monomial Factors 16 

30. Factoring by grouping Terms 17 

31. Factors of a Quadratic Trinomial 18 

32. Factoring the Difference of Squares 20 

33. Reduction to the Difference of Squares 20 

34. Replacing a Parenthesis by a Letter 21 

35. Factoring Binomials of the Form a" ± 6« 22 

36. Highest Common Factor 22 

S'7. H.C.F. of Two Polynomials 23 

i^.:)i}adid"s Method of finding the H.C.F 23 

39. Method of finding the H.C.F. of Two Polynomials ... 24 

40. Least y^o>nmon Multiple 26 

41. Second Rule for finding the Least Common Multiple ... 26 



CHAPTER III 
FRACTIONS 

42. General Principles . 27 

43. Principle I 27 

44. Principle II 27 

45. Principle III 27 

46. Reduction. . 27 

47. Least Common Denominators of Several Fractions ... 28 

48. Addition of Fractions 29 

49. Subtraction of Fractions 29 

60. Multiplication of Fractions 29 

51. Division of Fractions 29 



CHAPTER IV 
EQUATIONS 

52. Introduction . 32 

63. Identities and Equations of Condition 32 

64. Linear Equations in One Variable 33 

65. Solution of Problems . . 37 

66. Linear Equations in Two Variables 40 

57. Solution of a Pair of Equations 40 



CONTENTS 



vu 



SECTION 

68. Independent Equations 

69. Solution of a Pair of Simultaneous Linear Equations 

60. Incompatible Equations 

61. R^um6 

62. Solution of Problems involving Two Unknowns 

63. Solution of Linear Equations in Several Variables 



PAGE 
. 41 

• 42 
. 42 

43 
• 45 

47 



CHAPTER V 
RATIO AND PROPORTION 

64. Ratio 49 

65. Proportion . 49 

66. Theorems concerning Proportion 49 

67. Theorem v 50 

68. Mean Proportion 60 



CHAPTER VI 



IRRATIONAL NUMBERS AND RADICALS 



69. Existence of Irrational Numbers . 

70. The Practical Necessity for Irrational Numbers 

71. Extraction of Square Root of Polynomials . 

72. Extraction of Square Root of Numbers . 

73. Approximation of Irrational Numbers . 

74. Sequences 

75. Operations on Irrational Numbere 

76. Notation 

77. Other Irrational Numbers .... 

78. Reduction of a Radical to its Simplest Form 

79. Addition and Subtraction of Radicals . 

80. Multiplication and Division of Radicals . 

81. Rationalization 

82. Solution of Equations involving Radicals 



62 
53 
53 
54 
55 
56 
56 
57 
57 
68 
69 
60 
61 
63 



CHAPTER VII 
THEORY OF INDICES 

83. Negative Exponents 66 

84. Fractional Exponents 66 

85. Further Assumptions 67 

86. Theorem 67 

87. Operations with Radical Polynomials 69 



viii CONTENTS 



QUADRATICS AND BEYONI? 



i CHAPTER VIII 
QUADRATIC EQUATIONS 

SECTION PAGE 

88. Definition 70 

89. Solution of Quadratic Equations 70 

90. Pure Quadratics 72 

91. Solution of Quadratic Equations by Factoring .... 75 

92. Solution of an Equation by Factoring 75 

93. Quadratic Form 77 

94. Problems solvable by Quadratic Equations 79 

95. Theorems regarding Quadratic Equations 82 

96. Theorem 83 

97. Theorem 84 

98. Nature of the Roots of a Quadratic Equation 84 



CHAPTER IX 
GRAPHICAL REPRESENTATION 

99. Representation of Points on a Line 87 

100. Cartesian Coordinates 88 

101. The Graph of an Equation 90 

102. Restriction to Coordinates 91 

103. Plotting Equations 91 

104. Plotting Equations after Solution 93 

105. Graph of the Linear Equation . "94 

106. Method of plotting a Line from its Equation .... 96 

107. Solution of Linear Equations, and the Intersection of their Graphs 97 

108. Graphs of Dependent Equations 99 

109. Incompatible Equations 09 

110. Graph of the Quadratic Equation 100 

111. Form of the Graph of a Quadratic Equation .... 101 

112. The Special Quadratic ox^ + to = 103 

113. The Special Quadratic aa;^ + c = 104 

114. Degeneration of the Quadratic Equation 104 

115. Sum and Difference of Roots 106 

116. Variation in Sign of a Quadratic 107 



CONTENTS ix 



CHAPTER X 
SIMULTANEOUS QUADRATIC EQUATIONS IN TWO VARIABLES 

SECTION PAGE 

117. Solution of Simultaneous Quadratics Ill 

118. Solution by Substitution Ill 

119. Number of Solutions 113 

120. Solution when neither Equation is Linear 114 

121. Equivalence of Pairs of Equations 120 

122. Incompatible Equations 121 

123. Graphical Representation of Simultaneous Quadratic Equations 122 

124. Graphical Meaning of Homogeneous Equations . . . 123 

CHAPTER XI 
MATHEMATICAL INDUCTION 

126. General Statement 125 

CHAPTER XII 
BINOMIAL THEOREM 

126. Statement of the Binomial Theorem 128 

127. Proof of the Binomial Theorem 129 

128. General Term 129 

CHAPTER XIII 

ARITHMETICAL PROGRESSION 

129. Definitions : .... 133 

130. The nth Term 133 

131. The Sum of the Series 134 

132. Arithmetical Means 134 

CHAPTER XIV 

GEOMETRICAL PROGRESSION 

133. Definitions 137 

134. The nth Term 137 

135. The Sum of the Series 138 

136. Geometrical Means 138 

137. Infinite Series 140 



CONTENTS 



ADVANCED ALGEBEA 



CHAPTER XV 

PERMUTATIONS AND COMBINATIONS 
SECTION PAGE 

138. Introduction 143 

139. Permutations 144 

140. Combinations 146 

141. Circular Permutations 149 

142. Theorem 160 

CHAPTER XVI 

COMPLEX NUMBERS 

143. The Imaginary Unit 152 

144. Addition and Subtraction of Imaginary Numbers . . . 163" 

145. Multiplication and Division of Imaginaries 154 

146. Complex Numbers 156 

147. Graphical Representation of Complex Numbers .... 155 

148. Equality of Complex Numbers . 155 

149. Addition and Subtraction 156 

150. Graphical Representation of Addition 156 

151. Multiplication of Complex Numbers 157 

152. Conjugate Complex Numbers . . . . . . . 158 

153. Division of Complex Numbers 158 

154. Polar Representation 160 

155. Multiplication in Polar Form 160 

156. Powers of Numbers in Polar Form 161 

157. Division in Polar Form 162 

158. Roots of Complex Numbers 162 

CHAPTER XVII 



THEORY OF EQUATIONS 

159. Equation of the nth Degree 166 

160. Remainder Theorem 166 

161. Synthetic Division . . 167 

162. Proof of the Rule for Synthetic Division 169 

163. Plotting of Equations . . 170 

164. Extent of the Table of Values 171 



^ 



CONTENTS xi 

SIECTION PAGE 

fl65. Roots of an Equation 172 

. Number of Roots 172 

167. Graphical Interpretation 174 

^ 168. Imaginary Roots . . . 174 

. Graphical Interpretation of Imaginary Roots .... 175 

170. Relation between Roots and Coefficients 177 

171. The General Term in the Binomial Expansion .... 178 

172. Solution by Trial . 178 

173. Properties of Binomial Surds 179 

174. Formation of Equations . . 180 

176. To multiply the Roots by a Constant 183 

176. Descartes' Rule of Signs 186 

177. Negative Roots 189 

178. Integral Roots 190 

179. Rational Roots 190 

180. Diminishing the Roots of an Equation . . ... . 191 

181. Graphical Interpretation of Decreasing Roots .... 193 

182. Location Principle 194 

183. Approximate Calculation of Roots by Horner's Method . . 195 

184. Roots nearly Equal 200 



CHAPTER XVIII 

DETERMINANTS 

185. Solution of Two Linear Equations 203 

186. Solution of Three Linear Equations 204 

187. Inversion 208 

188. Development of the Determinant 208 

189. Number of Terms 210 

190. Development by Minors 210 

191. Multiplication by a Constant 213 

192. Interchange of Rows and Columns 213 

193. Interchange of Rows or Columns 214 

194. Identical Rows or Columns . . . . . . . 215 

195. Proof for Development by Minors 215 

196. Sum of Determinants 216 

197. Vanishing of a Determinant 217 

198. Evaluation by Factoring 218 

199. Practical Directions for evaluating Determinants . ... 219 

200. Solution of Linear Equations 221 ^ 

201. Solution of Homogeneous Linear Equations .... 223 4^^ — -• 



xii CONTENTS 

CHAPTER XIX 

PARTIAL FRACTIONS 
SECTION .^ PAGE 

202. Introduction 226 

203. Development when (f){x)= has no Multiple Roots . . . 225 

204. Development when (p{x)=0 has Imaginary Roots . . . 229 

205. Development when <p (x) = {x — ar)» 232 

206. General Case 233 

CHAPTER XX 
LOGARITHMS 

207. Generalized Powers 235 

208. Logarithms . ' 236 

209. Operations on Logarithms 237 

210. Common System of Logarithms 239 

211. Use of Tables 241 

212. Interpolation 242 

213. Antilogarithms 243 

214. Cologarithms 246 

215. Change of Base 247 

216. Exponential Equations 261 

217. Compound Interest 263 

CHAPTER XXI 
CONTINUED FRACTIONS 

218. Definitions . ' 256 

219. Terminating Continued Fractions 256 

220. Convergents 258 

221. Recurring Continued Fractions 260 

222. Expression of a Surd as a Recurring Continued Fraction . 263 

223. Properties of Convergents 266 

224. Limit of Error 267 

CHAPTER XXII 
INEQUALITIES 

225. General Theorems . .269 

226. Conditional Linear Inequalities 271 

227. Conditional Quadratic Inequalities 271 



CONTENTS xiii 



CHAPTER XXm 
VARIATION 

SECTION ~ PAOB 

228. General Principles . .273 



CHAPTER XXIV 
PROBABILITY 

229. Illustration 276 

230. General Statement 276 

CHAPTER XXV 
SCALES OF NOTATION 

231. General Statement 279 

232. Fundamental Operations ... .... 280 

233. Change of Scale ..... .... 281 

234. Fractions 282 

236. Duodecimals . , 284 



ik 



ADVANCED ALGEBRA 



ALGEBRA TO QUADRATICS 

CHAPTER I 
FUNDAMENTAL OPERATIONS 

1. It is assumed that the elementary operations and the mean- 
ing of the -usual symbols of algebra are familiar and do not 
demand detailed treatment. In the following brief exposition 
of the formal laws of algebra most of the proofs are omitted. 

2. Addition. The process of adding two positive integers a and 
h consists in finding a number x such that 

a -\-h = x. 

For any two given positive integers a single sum x exists 
which is itself a positive integer. 

3. Subtraction. The process of subtracting the positive num- 
ber h from the positive number a consists in finding a number x 

such that 

h -\- x = a. (1) 

This number x is called the difference between a and h and is 

denoted as follows : 

a — 6 = a;, 

a being called the minuend and h the subtrahend. 

If a > ft and both are positive integers, then a single posi- 
tive integer x exists which satisfies the condition expressed by 
equation (1) 

1 



. AL'G^BE a: to * ^JJ ADR ATI€S ^ 



li a <h, then x is not a positive integer. In order that the pro- 
cess of subtraction may be possible in this case also, we introduce 
negative numbers which we symbolize by (—a ), (— b), etc. When 
in the difference a~b,ais less than b, we define a — b = (— (b — a)). 
The processes of addition and subtraction for the negative numbers 
are defined as follows : 

a -^ (— b) = a — b. 

a — (— b) = a -^ b. 

4. Zero. If in equation (1), a = b, there is no positive or nega- 
tive number which satisfies the equation. In order that in this 
case also the equation may have a number satisfying it, we intro- 
duce the number zero which is symbolized by and defined by 

the equation 

a -\- = a, 

or a — a = 0. 

The processes of addition and subtraction for this new number 
zero are defined as follows, where a stands for either a positive or 
a negative number 

0-a = -a. 
± = 0. 

5. Multiplication. The process of multiplying a by i consists 
in finding a number x which satisfies the equation 

a-b = X. 

* The symbol for a positive integer might be written (+ a), (+ 6), etc., consistently 
with the notation for negative numbers. Since, however, no ambiguity results, we omit 
the + sign. Since the laws of combining the + and — signs given in this and the folloM'ing 
paragraphs remove the necessity for the parentheses in the notation for the negative 
number, we shall omit them where no ambiguity results. 



FUNDAMENTAL OPERATIONS 3 

When a and h are positive integers cc is a positive integer which, 
may be found by adding a to itseK h times. When the numbers 
to be multiplied are negative we have the following laws, 

(— a) • (— 5) = a • 5, 

Oa = «;.0 = 0, (1) 

where «; is a positive or negative number or zero. 

These symbolical statements include the statement of the 
following 

Principle. A 'product of numbers is zero when and only when 
one or more of the factors are zero. 

This most important fact, which we shall use continually, assures 
us that when we have a product of several numbers as 

a-b-G- d =^ Bj 

first, if e equals zero, it is certain that one or more of the num- 
bers a, bj c, or d are zero ; second, if one or more of the numbers 
a, b, c, or d are zero, then e is also zero. 

6. Division. The process of dividing ahj p consists in finding 
a number x which satisfies the equation 

X-p = a, . (1) 

where a and jS are positive or negative integers, or a is 0. 
When a occurs in the sequence of numbers 

• ••-.?A -2A -Ap, ft-2A 3A •••, 

a; is a definite integer or 0, that is, it is a number such as we 
have previously considered. If a is not found in this series, but 
is between two numbers of the series, then in order that in this 
case the process may also be possible we introduce the fraction 

which we symbolize by a -*- /8 or - and which is defined by the 
equation 



4 ALGEBRA TO QUADRATICS 

The operations for addition, subtraction, multiplication, and divi- 
sion of fractions are defined as follows : 

fi 8 Py W 

Further properties of fractions are the following : 

a 1 



— = — > where S is any number, (5) 



j8 SP 

The last two equations are expressed verbally as follows : 

Both numerator and denominator of a fraction may be multi- 
plied by any number without changing the value of the fraction. 

Changing the sign of either, numerator or denominator of a 
fraction is equivalent to changing the sign of the fraction. 

The laws of signs in multiplication given on p. 3 may now be 
assumed to hold when the letters represent fractions as well as 
integers. j- ,.-, . 

Thus for example ~ I t ) ' "~ 



ac 
M 



The positive or negative number a may be written in the 

fractional form 

a 

1* 

7. Division by zero. If in equation (1), ^Q>^ p=i 0, there is no 
single number x which satisfies the equation, since by (1), § 5, 
whatever value x might have, its product with zero would 
be zero. 



FUNDAMENTAL OPERATIONS 5 

Thus division by zero is entirely excluded from algebraic pro- 
cesses. Before a division can safely be performed one must be 
assured that the divisor cannot vanish. In the equation 

4 = 2.0, 

if we should allow division of both sides of our equation by zero, 
we should be led to the absurd result 4 = 2. 

8. Fundamental operations. The operations of addition, sub- 
traction, multiplication, and division we call the four fundamental 
operations. Any numbers that can be derived from the number 1 
by means of the four fundamental operations we call rational num- 
bers. Such numbers comprise all positive and negative integers 
and such fractions as have integers for numerator and denominator. 
Positive or negative integers are called integral numbers. 

9. Practical demaild for negative and fractional numbers. 
In the preceding discussion negative numbers and fractions have 
been introduced on account of the mathematical necessity for 
them. They were needed to make the four fundamental opera- 
tions always possible. That this mathematical necessity corre- 
sponds to a practiqal necessity appears as soon as we attempt 
to apply our four operations to practical affairs. Thus if on a 
certain day the temperature is + 20° and the next day the mer- 
cury falls 25°, in order to express the second temperature we 
must subtract 2^ from 20. If we had not introduced negative 
numbers, this would be impossible and our mathematics would 
be inapplicable to this and countless other everyday occurrences. 

10. Laws of operation. All the numbers which we use in 
algebra are subject to the following laws. 

' .Commutative law of addition. This law asserts that the value 
of the sum of two numbers does not depend on the order of 
summation. 

Symbolically expressed, 

a -\- h = h -\- aj 

where a and h represent any numbers such as we have presented 
or shall hereafter introduce. 



6 ALGEBRA TO QUADRATICS 

Associative law of addition. This law asserts that the sum of 
three numbers does not depend on the way in which the numbers 
are grouped in performing the process of addition. 

Symbolically expressed, 

a 4- (^ + c) = (a + &) + c = a 4- * + c. 

Commutative law of multiplication. This law asserts that the 
value of the product of two numbers does not depend on the order 
of multiplication. 

Symbolically expressed, 

a-b = b-a. 

Associative law of multiplication. This law asserts that the 
value of the product of three numbers does not depend on the way 
in which the numbers are grouped in the process of multiplication. 

Symbolically expressed, 

(^a-b) -c = a-(b'C)= a-b-c. 

Distributive law. This law asserts that the product of a single 
number and the sum of two numbers is identical with the sum 
of the products of the first number and the other two numbers 
taken singly. 

Symbolically expressed, 

a- (b -\- c)= a-b -{- a-c. 

All the above laws are readily seen to hold when more than three numbers are 
involved. 

11. Integral and rational expressions. A polynomial is inte- 
gral when it may be expressed by a succession of literal terms, no 
one of which contains any letter in the denominator. 

Thus 4 a;5 _ a;3 _ 2 x2 _ J x + 1 is integral. 

The quotient of two integral expressions is called rational. 

^^ a;2_2a; + 3. ^. , 
Thus z — 18 rational. 

a — 7 

12. Operations on polynomials. We assume that the same 
formal laws for the four fundamental operations enunciated in 
§ § 2-6 and the laws given in § 10 hold whether the letters in the 
symbolic statements represent numbers or polynomials. 



FUNDAMENTAL OPERATIONS 7 

In fact the literal expressions which we use are in essence 
nothing else than numerical expressions, since the letters are merely 
symbols for numbers. When the letters are replaced by numbers, 
the literal expressions reduce to numerical expressions for which 
the previous laws have been explicitly given. 

13. Addition of polynomials. For performing this operation 
we have the following 

EuLE. Write the terms with the same literal part in a column. 
Find the algebraic sum of the terms in each column, and write 
the results in succession with their proper signs. 

When the polynomials reduce to monomials the same rule is to be observed. 

EXERCISES 

Add the following : 

1. 3a262_2a6 + 6a26-a; 4. ab - 2 a^h'^ - 11 a'^b + 9 a ; 

2a^b -ab -2a; Sam- 4:db -6a. 

Solution: ZaW-2ab+ 6a^b- a 

-2a262 + 4a6- lla26 + 9a 

- a6 + 2 a26 - 2 a 

Sa^-iab -6a 

9am -Sab- Sa^b 

2. 21a-246-8c2; 16c + 17 6 + 6c2 - 20a ; 186 -18c. 

3. x* -6a;2- 8x-l; 2x3 + 1; 6x'^ + 1 x + 2 ; x^ - x^ + x - 1. 

4. 9(a + 6)-6(6 + c) + 7(a + c); 4(6 + c) - 7(a + 6) - 8(a + c) ,• 

(a + c) - (a + 6) + (6 + c). 

5. a2-4a6 + 62 + a + 6-2; 2a2 + 4a6- 362- 2a -26 + 4; 

3a2- 5a6-462 + 3a + 36 - 2; 6a2 + 10a6 + 562 + a + 6. 

14. Subtraction of polynomials. For performing this opera- 
tion we have the following 

EuLE. Write the subtrahend under the minuend so that terms 
with the same literal part are in the same column. 

To each term of the minuend add the corresponding term of 
the subtrahend, the sign of the latter having been changed. 

It is generally preferable to imagine the signs of the subtrahend changed rather 
than actually to write it with the changed signs. 



8 ALGEBRA TO QUADRATICS 

EXERCISES 

1. From a262 - 3 a^ft + 8 a6 + 6 6 subtract 9 a'^h'^ - 6 a6 + 4 a^ft + a. 

Solution : a%'^ - 3 a26 + 8 a& + 6 6 

-9a262:f 4a26- 6a6 +a 

- 8a262 ^ 7 a26 + 14a6 + 66 - a 

2. From 6 ahx — 4 win + 5 x subtract 3 mn + 6 ax — 4 a6x. 

3. From m -{■ an + bq subtract the sum of 

cm + dn + {b — a)q and {a — b)q — {a -\- d)n — cm. 

4. From the sum of | a + y^ 6 + | c and — 6 — c — a subtract ^6 — |c + ^a. 

5. From the sum of 2 x^ — 3 x + 4 and x* — f x — i subtract x^ — | x^ 

-3|x + 3i. 

15. Parentheses. When it is desirable to consider as a single 
symbol an expression involving several numbers or symbols for 
numbers, the expression is inclosed in a parenthesis. This paren- 
thesis may then be used in operations as if it were a single number 
or symbol, as in fact it is, excepting that the operations inside 
the parenthesis may not yet have been carried out. 

EuLE. When a single parenthesis is preceded hy a -\- sign 
the parenthesis may he removed, the various terms retaining 
the same sign. 

When a single parenthesis is preceded hy a — sign the parent- 
thesis may be removed, providing we change the signs of all the 
terms inside the parenthesis. 

When several parentheses occur in an expression we have the 
following 

EuLE. Remove the innermost parenthesis, changing the signs 
of the terms inside if the sign preceding it is minus. 

Simplify, if possible, the expression inside the new inn^rwjost 
parenthesis. 

Repeat the process until all the parentheses are removed. 

It is in general unwise to shorten the process by carrying out some of the steps 
in one's head. The liability to error in such attempts more than offsets the gain 
in time. 



FUNDAMENTAL OPERATIONS 



EXERCISES 

Remove parentheses from the following : 



1. 6 _ pa -[26 + (4a -2a -6) -66]}. 



Solution: 6 - {9a - [26 + (4a - 2a - 6) - 66]} 

= 6- {9a -[26 + (4a -2a + 6) -66]} 
= 6-{9a-[2 6+ (2a + 6)-6 6]} 
= 6 _ [9 a - (2 6 + 2 a + 6 - 6 6)] 
= 6- [9a -(2a -36)] 
= 6 -(9a -2a + 36) 
= 6 - (7 a + 3 6) 
= 6-7a-36 
= - 7 a - 2 6. 

2- -{-[-[-(-(-I))]]}- 

3. a2 + 4 - {6 - [- (a2 - 6) + 1]}. 

4. i{i-|[f-ia-i-f-T\) + |]-^}- 

5. x2 - {2/2 _ [4x + 3(y - 9x(y - x)) + 9y (x - y)]}. 

6. Find the value of a - {5 6 - [a - (3 c - 3 6) + 2 c - 3 (a - 26 -c)]} i^ 
when a = — 3, 6 = 4, c = — 5. 

16. Multiplication. It is customary to write a - a = a^ ; 
aaa = a^] a - a • • a = a"*. We have tlien by tlie associative 

n terms 

law of multiplication, § 10, 

a^ - a^ = (a • a) (a - a ' a) = a a • a ' a • a = a^, 
or, in general. 



where m and n are positive integers. 
Furthermore, 



(I) 

(a')" = a"-a'---a' = a'-"'. (II) 



m terms 



Finally, a" • J" = (a • bf. (Ill) 

The distinction between (a**)"* and a""* should be noted carefully. Thus (28)* 
= 82 = 64, while 28* =2^ = 512. 

Equation (I) asserts that the exponent in the product of two 
powers of any expression is the sum of the exponents of the 
factors. Hence we may multiply monomials as follows : 



10 ALGEBRA TO QUADRATICS 



I 



Rule. Write the product of the numerical coefficients, followed , 
by all the letters that occur in the multiplier and multiplicand, 
each having as its exponent the sum of the exponents of that 
letter in the multiplier and multiplicand. 

Example. 4 a2^i0c# • (- 16 a'^bd'^) = - 64 a^b^^cd^\ 

17. Multiplication of monomials by polynomials. By the 

distributive law, § 10, we can immediately formulate the 

EuLE. Multiply each term of the polynomial by the monomial 
and write the resulting terms in succession. 

Example. 9 a'^h'^ - 2 a& + 4 a62 - a + &* 

3a26 

27 a463 _ 6 aW + 12 a^h^ - 3 a^fe + 3 aW 
* 

18. Multiplication of polynomials. If in the expression for the 

distributive law, a(c + d)= ac •{- ad, 

we replace a by a + 5, we have 

(a -\- h) (c -\- d) = ac -\- be -\- ad -{- bd, 
which affords the 

EuLE. Multiply the mult^icand by each term of the multi- 
plier in turn, and write the partial products in succession. 

To test the accuracy of the result assume some convenient numer- 
ical value for each letter, and find the corresponding numerical 
value of multiplier, multiplicand, and product. The latter should 
be the product of the two former. 

EXERCISES 



1. Multiply and check the following: 




(a) 2 a2 + a6 + 4 62 + 5 and a-h-\- ah. 


Check : 


Solution : 


Let a = 6 = 1 


2a2+ a6+462 + 6 


= 8 


a - h + a6 


= 1 


2a8+ a26 + 4a&2 + a6 




-2a26- a62 -463_ 


-62 


+ a62 


+ 2a86 + a262 + 4a68 



2a8 - a26 + 4a62 + a6 - 46^ - 62 -f- 2a86 + a262 + 4a68 = 8 



FUNDAMENTAL OPERATIONS H 

(b) 6 abx^ and 4 a'^b^x. 

(c) ^^and -OxV^. 

o 

(d) 3 a62x - ^ 6x4 and 6 a^cx. 

(e) x2« + ?/26 ^ x^ and x« — y^ 

(f ) a^ + ab + l^ and a^ + ac + c2. 

(g) x^y^'^j a;»-3yin4-4^ and ic^y2m-2^ 
(h) xP-3 4- xP-2 + 1 and a;3 _ a;2 _ 1, 

(i) 8 a26c, - a62, _ 7 62, - — a^c*, and -. 
w ' 4 ' ' 14 ' 6 

(j) ax* — 2 a2x3 — X + 4 a and — x + 2 a. 
(k) x« + * + x2« + x2& 4- a;3o-& and x«-^ — 1. 

(1) 15x* - 11x3 + 6x2 + 2x - 1 and - 3x2 - 1. 
(m) 4 x* — 8 xy^ + | x2y2 _ 3 x^y — x — y and — 42 xy. 

2. Expand (x + y)*. 

3. Expand and simplify 

(a;2 + 2/2 + 22)2 _ (X + y + 2r) (X + y - z) (x + 2 - y) (y + 2 - X). 

19. Types of multiplication. The following types of multi- 
plication should be so familiar as merely to require inspection of 
the factors in order to write the product. 

EuLE. The product of the sum and difference of two terms is 
equal to the square of the terms with like signs minus the square 
of the terms which have unlike signs. 

Examples. (a - 6) (a + 6) = a?^ — 62. 

(4x2 - 3y2)(4x2 + 3y2) = 16x4 - 9y4. 

20. The square of a binomial. This process is performed as 
follows : 

KuLE. The square of a binomial, or expression in two terms, 
is equal to the sum of the squares of the two terms plus twice 
their product. 

Examples. (x + y)2 = x2 + y2 + 2 xy. 

(2a-36)2=:4a2 + 962- 12a6. 



12 ALGEBRA TO QUADRATICS 

21. The square of a polynomial. This process is performe! 
as follows : 

Rule. The square of any polynomial is equal to the sum 
of the squares of the terms plus twice the product of each term 
hy each term that follows it in the polynomial. 

Example, (a + 6 + c)2 = a^ + 62 + c^ + 2 a6 + 2 ac + 2 6c. 

22. The cube of a binomial. This process is performed as 
follows : 

Rule. The cube of any binomial is given hy the folloimng 

expression : ^^ _^ ^y ^ ^s _^ ^ ^2j -\.3ab^ + h\ 

EXERCISES 

Perform the following processes by inspection. 
1. (a-6 + c)2. 2. (a4-66)2. 

3. (2x'-i-l)2 4. (a2-62)3. 

5. (2x'-i-l)3. 6. (l-8a;22/)2. 

7. (X2-2X + 1)2. 8. (a;2 - y2 + ^2)2. 

9. (2a-26-c)2. 10. (x8-2x-l)2. 

11. (ai'-3-6p + 3)2. 12. (-6x2y + 4xy2)8. 

13. {xp - 2/9) {xp + y^). 14. (- 6x22/ + 4 xy^)^. 

15. (3x+ 2 2/)(3x-2 2/). 16. (- 3ax2 + 2ax -6)2. 

17. {-3x^y-^lz^){Sx'^ + ^z^). 18. (_ 4 - 6a26) (- 4 + 6a26). 

19. (2 a -2^. 20. (2 a-?)'. 

23. Division. By the definition of division in § 6, we have 

a = —} a'^ = — ) a^ = —z'y 
. a a a'' 

or, in general, ^ 

a"-™ = — -> 
a"" 

where n and m are positive integers and n> m. 

If n = m, we preserve the same principle and write 

a«-» = a« = — = 1, 



(1) 



FUNDAMENTAL OPERATIONS 13 

24. Division of monomials. Keeping in mind the rule of signs 
for division given in § 6, we have the following 

EuLE. Divide the numerical coefficient of the dividend hy 
that of the divisor for the numerical coefficient of the quotient^ 
keeping in mind the rule of signs for division. 

Write the literal part of the dividend over that of the 
divisor in the form of a fraction^ and perform on each pair 
of letters occurring in both numerator and denominator the 
process of division as defined hy equation (I) in the preceding 
paragraph. 

Example. Divide 12 a'^lA^cH by - 6 a^hc^d^. 
12 a^6"c2d _ 2 bio 

25. Division of a polynomial by a monomial. This process is 
performed as follows : 

Rule. Divide each term of the polynomial hy the monomial 
and write the partial quotients in succession. 
Example. Divide 8 a^lP - 12 a^h^ by 2 aW. 

8a266 _ 12 gcftg _ 463 _ 6a» 

2a363 2a363 ~ a & * 

26. Division of a polynomial by a polynomial. This process 
is performed as follows : 

Rule. Arrange hoth dividend and divisor in descending 
powers of some common letter {called the letter of arrangement)- 

Divide the first term of the dividend hy the first term of the 
divisor for the first term of the quotient. 

Multiply the divisor hy this first term of the quotient and 
subtract the product from the dividend. 

Divide the first term of this remainder hy the first term of the 
divisor for the second term of the quotient, and proceed as 
before until the remainder vanishes or is of lower degree in the 
letter of arrangement than the divisor. 



14 ALGEBRA TO QUADRATICS 

When the last remainder vanishes the dividend is exactly divisible by 
divisor. This fact may be expressed as follows: 

dividend _ . . . 

,. . = quotient. 

divisor 

When the last remainder does not vanish we may express the result of division 

dividend .. , , remainder 

-^rr-. = quotient -\ -— : 

divisor divisor 

The coefficients in the quotient will be rational numbers if those in both divi- 
dend and divisor are rational. 

EXERCISES 

Divide and check the following : 

1. 8a8 + 6a26 + 9a62 + 963by4a + &. 

Solution : 4a4-&| 8 a^ + 6 a26 + 9 a62 + 9 h^ \2 a^ + ab + 2lfl 

8 a3 + 2 a'^b 

4 a26 + 9 a62 
4a26+ am 

8a&2 + 9&3 
8a&2 + 268 
76» 
7 63 . ■ 

Result : 2a^-\-ah-{-2,b^ + 

1 4a + 6 • 

Check : Let a = h =1. Dividend = 32, divisor = 5, quotient = 6§. 
32 - 5 = 6f . 

2. xi2 - yi2 by x3 - 2/3. 3. 2 x2 - 6x + 2 by X - 2. 
4. xi2 — 2/12 i3y aj4 _ 2/*. 5. x^ — 2/6 by x^ + xy -\- y^. 
6. a8 - a2 + 2 bya + 1. 7. - 63x*y32;2 ^y _ gx^y^z. 

8. .x2 - X - 30 by X + 5. 9. 4 a26 - 6 a62 _ 2 a by - 2a. • 

10. 16 a264cii by - 2 a^¥c\ 11. i x2 - 3^ x - | by 1^ x + T»ff. 

12. ax2 + (a2 - 6) X - a6 by X + a. 

13. ax« + 6x»-i + cx«-2 - dx«-3 by x'. 

14. 16 a2x22/2 _ 8 ax32/2 - 4 x*?/ by - f xy. 

15. ISai'b^ -\- 6aP + 2?^+3_ 9 ap + ^ft^ by 3 apft*. 

16. a2 - 2 a6 - 4 c2 + 8 6c - 3 62 by a - 2 c + 6. 

17. x* — (d + 6 + c) x2 -f {ab + ac + 6c) x - a6c by x - a. 

18. 2j/2 _ 6x2 + i/xy + V-x - -^y + 1 by 2x + |y - f. 

19. x8 - 2 x22/ - x2 + 2/2x f 2 xy - y - 2/2 + 2 by X - y + 1. 

20. 3x8 + 6x22/ + 9x2 + 2x2/2 + 5^3 + 22/ + 6y2 + 3 by x + 22/ + 3. 



FUNDAMENTAL OPERATIONS 15 

27. Types of division. The following types of division, which 
may be verified by the rule just given for any particular integral 
value of rij should be so familiar that they may be performed by 
inspection. 

(a^n _ ^,2n) ^ (^^» -t ^n^ = «»» ^ ^«. (I) 

(a" + b^) -^ (a + 5) = a"-i - a^-% + a^-%'' + b^-\ (II) 

where n is odd. 

(a" -. h^) ^(a-b)^ a"-i 4- a"-'^ + a«-3^>2 + . . . + j«-i^ (III) 
where n is odd or even. 

EXERCISES 

Give by inspection the results of the following divisions. 
1. a« - 1 by a -r l: 2. a^ + 1 by a + 1. 

3. x' + 128 by ic + 2. 4. x^ + y^hy x-\- y. 

6. x^ — y^ by x^ -{■ y^. 6. x^ — y* hj x — y. 

7. x8 - ^8 by X* - y*. 8. a2m _ 1 by a - 1. 

9. a2«+^ - 1 by a - 1. 10. 27a9 + 868 by 3^8 + 26. 

11. 8x8-27 by 2a:- 3. 12. 4a2 - 25668 by 2a + 1664. 

13. 16 a* - 256 by 4 a^ + 16. 14. 27 ai^ - 64 612 by 3 a* - 4 6*. 



CHAPTER II 
FACTORING 

28. Statement of the problem. The operation of division con- 
sists in finding the quotient when the dividend and divisor are 
given. The product of the quotient and the divisor is the divi- 
dend, and the quotient and the divisor are the factors of the 
dividend. Thus the process of division consists in finding a 
second factor of a given expression when one factor is given. 

The process of factoring consists in finding all the factors of 
a polynomial when no one of them is given. This operation is in 
essence the reverse of the operation of multiplication. We shall 
be concerned only with those factors that have rational coefficients. 

29. Monomial factors. By the distributive law, § 10, 

ah -{• ac = a(}) -\- c). 
This affords immediately the 

Rule. Write the largest monomial factor which occurs in every 
term outside a jpareiithesis which includes the algebraic sum of 
the remaining factors of the various terms. 



EXERCISES 

Factor the following : 

1. 6 Q?Wc + 9 alPc^ - 15 a*6c7. 

Solution : 6 a'^hH -t- 9 ah^c^ - 15 a^hc^ = 3 a6c (2 at^ -J- 3 ftScS - 5 a^<fi). 

2. 14 anx — 21 6nic — 7 n. 

3. 121 a'^hH - 22 a^hc'^ + H oJb'^cK 

4. 5xV - 10x8?/8 - SxV - 15x2y2. 

5. 21 ahn + 6 obH"^ - 18 a'^hv?- + 15 a'^h'^n. 

6. 10 al^cmx - 5 ab^cy + 6 ab'^cz - 15 abc^m^. 
7 a^xV - 49 ax^y* + 14 axy^z^ - 21 a^'^y^. 

8. 45 a*62c8d - 9a6*c«d8 -I- 27 a^bc*d^ - 117 a^l^cd*. 

16 



FACTORING 17 

30. Factoring by grouping terms. If in the expression for 
the distributive law, ac + bc = {a-\- b) c, 

we replace chj c -\- d, 

we have a(G -\- d) -\- h (c -\- d) = ac -\- ad -\- be -\- bd. 

We may then factor the right-hand member as follows : 
ac_ -^ ad -\- be -\- bd = a (c -\- d) -\- b {c -\- d) = {a -{- b) (c -\- d). 

This affords the 

EuLE. Factor out any monomial expression that is common to 
each term of the polynomial. 

Arrange the terms of the polynomial to be factored in groups 
of two or more terms each, such that in each group a monomial 
factor may be taken outside a parenthesis which in each case 
contains the same expression. 

Write the algebraic sum of the monomial factors that occur 
outside the various parentheses for one factor, and the expres- 
sion inside the parentheses for the other factor. 

EXERCISES 

Factor the following : 

1. 4a3&-6a262_4a4 + 6a63. 

Solution : 4 a^b - 6 a'^h'^ - 4 a* + 6 aft* 

= 2 a (2 a26 - 3 a62 _ 2 a3 + 3 63) 

= 2 a (2 a26 - 2 a^ - 3 a62 + 3 &») 
= 2 a [2 a2 (6 ^a) + 3 62 (6 _ a)] 
= 2 a (6 - a) (2 a2 + 3 62). 

2. x2-(3a + 46)x + 12a6. 

Solution : x2 - (3 a + 4 6) x + 12 a6 

= x2 - 3 ax - 4 6x + 12 a6 
= X (x — 3 a) — 4 6 (x — 3 a) 
= (x-46)(x-3a). 

3. 2 ax - 3 6y - 2 6x + 3 ay. 4. 56 a2 - 40 a6 + 63 ac - 45 6c. 

5. a''x2 — 6'?y3 _ f/jx^ + apy\ 6. 91 x2 - 112 mx + 65 nx - 80 mru 

7. ax - bx -[■ ex + ay - by -\- cy. 8. 2 ax - 6x - c6 + 2 a6 + 2 ac - 62. 



18 ALGEBRA TO QUADRATICS 

9. 2x6 - 3x8 + 2x« - 3. 10. x^ + x^ + x + 1. 

11. acx"^ - hex + adx-bd. 12. 2 6x - x^ - 4 & + 2 x. 

13. 3x3 _ 12x82/2 _ 4y2 + 1. 14. 4x2 - (8 + b)x + 2 6. 

15. x* - (4m + 9n)x2 + 36 mn. 16. x* - (2m + 3n)x2 + 6mn. 

17. a2x-ac+a6y-a62x-63y +c62. 18. 18 a» - 2 ac*&6 _ g a26 + c^b^. 
19. 2ax — 6ay + a-26x + 56y-6. 20. 2ax-ay-26x + 4cx— 2cy+62/. 

21. 2 a2x« + 4 a2x* + 2 a2x2 + 4 a^. 

22. 8x2 - 2 ox - 12 xz + 3az + 4xy - ay. 

31. Factors of a quadratic trinomial. In this case we cannot 
factor by grouping terms immediately, as that method is inappli- 
cable to a polynomial of less than four terms. We observe, how- 
ever, that in the product of two binomial expressions. 

(mx + n) (px + qy= tnp^J -h {mq + w^)ic -|- ngy 

the coefficient of x is the sum of two expressions mq and np^ 

whose product is equal to the product of the coefficient of x^ and 

the last term, that is, 

mq • np = m,p • nq. 

Thus, to factor the right-hand member of this equation, we may 
remove the parenthesis from the term in x and use the principle 
rf grouping terms. Thus 

mpx^ -f (mq + np) q? + nq 

= TYipx^ -j- m.qx + npx -f- nq 
= mx (px -\- q)+ n (px -f q) 
= (mx + n) (px -\- q). 
This affords the 

Rule. Write the trinomial in order of descending powers of 
X (or the letter in which the expression is quadratic). 

Multiply the coefficient of a^ by the term not involving x, arid 
find two factors of this product whose algebraic sum is the 
coefficient of x. 

Replace the coefficient of x by this sum and factor by group- 
ing terms. 



FACTORING 19 

Factoring a perfect square is evidently a special case under this method. 
Thus factor x2 + 6x + 9. 

1-9=9. 3 + 3 = 6. 

x2 + (3 + 3)x + 9 
= x2 + 3x + 3x + 9 
= a;(x + 3) + 3(x + 3) 
= (X + §) (X + 3) 

= (X + 3)2. 

One will usually recognize when a trinomial is a perfect square, in which 
case the factors may be written down by inspection. 



+ 2 = 8. 







EXERCISES 


Factor the following : 


( 


1. 3x2 + 8x + 4. 


/ 


Solution : 


3-4 = 


12. / 6 H 






3x2 + 8x + 4 
= 3x2 + (6 + 2)x + 4 
= 3x2 + 6x + 2x + 4 
= 3x(x + 2) + 2(x + 2) 
= (3x + 2)(x + 2). 


2. 8x2- 


146X + 352. 




Solution : 


8. 362:.. 2462. 






8x2-146x + 362 



26 -.126 = -146. 

= 8x2-(2 6 + 12 6)x + 362 
= 8x2-126x-26x + 362 
= 4x(2x-36)-6(2x-36) 
= (4x-6)(2x-36). 
3. 28x2 -3x- 40. 

Solution : 28 • (- 40) = - 1120. 

The factors of 1120 must be factors of 28 and 40. We seek two factors 
of 1120, one of which exceeds the other by 3. We note that since 40 exceeds 
28 by more than 3, one factor must be greater and the other less than 28 and 
40 respectively. 

Since 4 • 7 = 28 and 5 ■ 8 = 40, 

we try 5 • 7= 35 and 4 • 8 = 82, 

which are the required factors of 1120. 

28x2 -3x- 40 
= 28x2-(35-32)x-40 
= 28 x« - 35 X + 32 X - 40 
= 7x(4x-6) + 8(4x-6) 
= (7x + 8)(4x-5). 



20 ALGEBRA TO QUADRATICS 

4. x2-6x + 9. 5. 2x2 + re -6. 

6. 2x2 - X - 6. 7. 2x2 + X - 91. 

8. x2 + X - 182. 9. 9x2 - 2x - 7. 

10. 2x2 + 5a; + 3. 11. x2 - llx + 18. 

12. 3x2-10x-8. 13. 6x2 + 17 x + 7. 

14. 16x2 + 4x- 3. 15. 7?/2-4?/-ll. 

16. a2-6a6 + 962. 17. 5x2-12x + 4. 

18. 18x2-73x + 4. 19. 27x2 + 3x- 2. 

20. 24x2 - 31x - 16. 21. 21x2 - 31x + 4. 

22. 12 x2 + 60 X - 72. 23. x* - 3 az^-A 2 a^. 

24. 9x2-18ax-7a2. 25. 10x2-63x-13. 

26. x4y« - 12 x2?/3 + 36. 27. 4x2p-16xi'- 81. >v 

28. 2 x8 - 17 6x2 + sif'^x. 29. 4 a2 + 12 a6 + 9 62. 

30. 6 a2x2 - 2 a6x - 7 62. ' 3 1. 10 x* - 16 a2x3 - 100 x^a^. 

32. 4 aa"* + 16 a^b» + 16 62«. 33. 4 a^x^y* - 20 a6xy2z + 25 62z2. 

32. Factoring the difference of squares. Under the method of 
the preceding paragraph we may factor the difference of squares. 

Thus to factor x^ — b^ we observe that the product of the 
coefficient of x^ and the constant term is 

1 . (_ b^) = _ b^. 

Since the coefficient of x. is zero, we have 

-b-\-b = 0. 

Hence (x -\- b)(x — b)— x^ — b\ 

EuLE. Extract the square root of each term. 
The sum of these square roots is one factor, and their differ- 
ence is the other. 

Example. Factor 9a2xV - 16 68c2. 

9 a2x«2/* - 16 68c2 = (3 axV + 4 64c) (3 ax-3y2 _ 4 54c) . 

33. Reduction to the difference of squares. The preceding 
method may be used when the expression to be factored becomes 
a perfect square by the addition of the square of some expression. 



FACTORING 21 

EXERCISES 

Factor the following : 

1.44 ^4 6*. 

Solution r a* + 4 6* = a^ + 4 0,262 + 4 54 _ 4 ^^252 

= (a2 + 2b2)2_4a262 
= (a2 + 2 62 _ 2 ab) {a^'-\- 2 62 + 2 db). 

2. 1 - a*. - 3. a* + 4. 

4. x^ — X. 5. x«y4 + 4 x2. 

6. 4a;4+.2/*. 7. 4a2-2562. 

8. x* + ic2 ^. 1. 9 ig ^254 _ 3.4. 

10. x4 + 9x2 + 81. 11. 4a2p - 962c2«. 

12. a2p + 3 - 16a36*. 13. ar*« + x2« + 1. 

14. 36x2^4^8 - 49 w2ui6. 15. x4 - 13x2 + 36. 

16. nty^ X IG mx* - lij m'x^y'^. 1 7. 9 x* + 8 x^^ + 4 y^. 

34. Replacing a parenthesis by a letter. Any of the preceding methods 
may be applied when a polynomial appeare in place of a letter in the expres- 
sion to be factored. It is frequently desirable for simplicity to replace such 
a polynomial by a letter, and in the final result to restore the polynomial. 

EXERCISES 

Factor the following : 

1. 2 ax2 - 2 6x2 - 6 ax + 6 6x - 8 a + 8 6. 

Solution : 2 ax2 - 2 6x2 - 6 ax + 6 6x - 8 a + 8 6 

= 2(a — 6)x2 - 6(a - 6)x - 8(a - 6) 
= (a -6) (2x2 -6.x -8) 
= 2{a-6)(x2-3x-4) 
= 2 (a - 6) (X - 4) (X + 1). 

In this example the factor (a — 6) might have been replaced by a letter. 

2. a2 + 62 - c2 - 9 - 2 a6 + 6 c. 

Solution : a2 + 62 - c2 - 9 - 2 a6 + 6c 

= a2 - 2 a6 + 62 - (c? - 6c + .9) 
= (a - 6)2 -^ (c - 3)2 . 

= (a - 6 + c - 3) (a - 6 - c + g. 

3. (3x-2/)(2a+p)-(3x-2/)(a-9). 

4. (4a-66)(3m-2p) + (a + 56)(3m-2p). 

5. (7a-32/)(5c-2d)-(6a-22/)(5c-2d). 

6. (X - 2/) (3a + 46) - (4a - 56) (X - 2/) - (X - y) (2a - 86). 



22 ALGEBRA TO QUADRATICS 

7. 6(x + 2/)2-ll(x4-2/)-7. 

8. 4a2 - 12a6 + 962 _ x2 - 2x - 1. 

9. x2a2 + 2 x2a + x2 - a2 - 2 a - 1. 
10. ax2 + 6 ox + 9a - &x2 - 66x - 96. 

fll. 4 (a - 6)2 - 5(a2 - 62) - 21 (a + 6)2. 

12. 6(x + ?/)2 - 12 (x2 -y2)^4{x- yf. 

13. a262x2 - a262 - 2 a6x2 + 2 a6 + x2 - 1. 

14. (x-2y)(2a-36)-(96-10)(x-2?/). 

/ 35. Factoring binomials of the form d^ ± 6*^. By § 27, 

b/ ^^ V ci" + ^" = (« + ^'Xa"-^ - a"-2^ + a«-3^»2 _|_ ^,«-i>)^ 

where w is odd. 

One can factor by inspection any binomial of the given form 
by reference to these equations. 

EXERCISES 

Factor the following : 

1. x^ — 2/6. 

Solution : x^ - ye - ^xB _ ^3) (a;3 + ^3) 

= (X2 + Xy + 7/2) (X - y) (X2 - Xy + 2/2) (x _|. y). 

2. x6 + 125. 3. x^ - 1. 
4. xi2 - 2/12. 5. x^ — 2/9. 

6. Xl8 - 2/18. 7. a;16 _ yl6. 

8. a2x3 + a5. 9. x* - a82/4. 

. 10. 216 a + a*. 11. ox* - 16 a. 

12. 3a7 - 96 65a2. 1^-"" 13. 2,1 x^^ + 64 2/8. ^"^^ 

14. 27 x52/7 + x22/4. «--^ 15. 16a468-81ci6d8,/^ 

36. Highest common factor. An expression that is not further 
divisible into factors with rational coefficients is called prime. 

If two polynomials have the same expression as a factor, this 
expression is said to be their common factor. 

The product of the common prime factors of two polynomials 
is called their highest common factor, or H.C.F. 

The same common prime factor may occur more than once. Thus (x — 1) 2 (x + 1) 
and (x - 1)2 (a; - 2)2 have (x -X^ as their H.C.F. v-^— 



c^f 



FACTORING 23 

37. H.C.F. of two polynomials. The process of finding the 
H.C.F. is performed as follows: 

EuLE. Factor the polynomials. The product of the common 
prime factors is their H.C.F. 

EXERCISES [l^'^-Xi^\)i^^^UH 

Find the H.C.F. of the following: 

1. 4 ab^x^ - 8 ab^x^ + 4 ab^ and 6 abx"^ + 12 abx + 6ab. 
Solution : — 4 ab'^x^ - 8 ab^^ + 4 ab^ 

= 4a62(x4- 2x2+1) 
= 4a62(x-l)2(x + l)2. 

6 a6x2 + 12 abx + 6ab 
= 6a6(x2 + 2x + 1) 
= 6a6(x + 1)2. 

The H.C.F. is then 2 a6(x + 1)2. 

2. x^ — y^ and x2 — y"^. 

3. x3 + x2 - 12 X and x2 + 5x + 4. 

4. 9mx2 — Qmx + m and 9nx2 — n. 

5. 6x- 4x2 + 2 ax -3a and 9 - 4x2. 

6. 12 a2 - 3(3a6 + 27 62 and 8a2 - 1862. 

7. 3 a2x - 6 a6x + 3 62x and 4 02^/ - 4 62y. 

8. 2 X - 46 - x2 - 2 6x and 4x - 5x2 - 6. 

9. 6x^-7 ax2 - 20 a2x and 3 x2 + ax - 4 a2. 

38. Euclid's method of finding the H.C.F. When one is unable to factor 
the polynomials whose H.C.F. is sought, the problem may nevertheless be 
solved by use of a method which in essence dates from Euclid (300 e.g.). 

The validity of this process depends on the following 
Principle. If a polynomial has a certain factor, any multiple of it has the 
same factor. 

Let x» + ^x»-i + 5x»-2 + . . . +£: 

and . x« + ax'»-i +6x'»-2 H 1- ^ 

be represented by F and 6? respectively. The letters A, B, ■ ■, K and a, 6, 

• • •, Z represent integers, and m, the degree of G, is no greater than n, the 

degree of F. We seek a method of finding the H.C.F. of F and G if any 

exists. Call Q the quotient obtained by dividing F by G, and call B the 

remainder. Then (§ 26) 

^' F=QG-^B, (1) 



24 ALGEBRA TO QUADRATICS 

where the degree of E in x is not so great as that of G. Now whatever the 
H.C.F. of i^'and G may be, it must also be the H.C.F. of G and B. For since 

F-QG = B, 

the H.C.F. of F and G must be a factor of the left-hand member, and hence 
a factor of R, which is equal to that member. Also every factor common 
to G and R must be contained in F, for any factor of G and E is a factor of 
the right-hand member of (1), and hence of F. 

Thus our problem is reduced to finding the H. C. F. of G and R. Let Qi and 
Bi be respectively the quotient and remainder obtained in dividing G by R. 

Then (? = QiB 4- Ru 

where the degree of Ei in x is not as great as that of B. By reasonirg simi- 
lar to that just employed we see that the H.C.F. of G and R is also the 
H.C.F. of B and Bi. Continue this process of division. 

Let B = Q2R1 + R2, 

-Bi = QsRi + Eg. 
until, say in Rk = Qk+2Rk + \ + Bk+i, 

either B^ is exactly divisible by B^ +1 (i.e. E^- + 2 = 0), or i?;t + 2 does not con- 
tain X. This alternative must arise since the degrees in x of the successive 
remainders E, Ei, E2, • • • are continually diminishing, and hence either the 
remainder must finally vanish or cease to contain x. Suppose Bk + 2=0. Then 
the H.C.F. of Bk and Bk + 1 is Ea.- + 1 itself, which must, by the reasoning given 
above, be also the H.C.F. of F and G. If Rk + 2 does not contain x, then the 
H.C.F. of F and G, which must also be a factor of Ea + 2, can contain no x, 
and must therefore be a constant. 

Thus F and G have no common factor involving x. 

This process is valid if the coefficients of F and G are rational expressions in 
any letters other than x. 

39. Method of finding the H.C.F. of two polynomials. The above dis- 
cussion we may express in the following 

Rule. Divide the polynomial of higher degree {if the degrees of the polyno- 
mials are unequal) by the other, and if there is a remainder, divide the divisor 
by it ; if there is a remainder in this process, divide the previous remainder 
by it, and so on until either there is no remainder or it does not contain the letter 
of arrangement. If tJiere is no remainder in the last division, the last divisor is 
the H. C. F. If the last remainder does not contain the letter of arrangement, 
then the polynomials have no common factor involving that letter. 

In the application of this rule any divisor or remainder may be multiplied or 
divided by any expression not involving the letter of arrangement without affect- 
ing the H.C.F. 



FACTORING 



?6 



EXERCISES 
Find the H.C.F. of the following: 

2x 



1. 2x* + 2x3-x2 
Solution : /Q ^ 

Multiply by - 



1 andx4 + x3 + 4x + 4. 

2J2x* + 2x3-x2- 2x-l |xH x^+ 4x + 4 

2x^ + 2x3 + 8x + 8 

■1, -x2-10x-9 *■ ^' ^ 

x^-\-10x+9\ x^+ x3+ 4x + 4 |x2-9x + 81 



Ri 



x4 + 10x3+ 9x2 



,- [^-«) 



- 9x3- 9x2+ 4x ^t4H/l 

- 9x3-^90x2- 81x I 



Divide by - 725, 



(9-!. c , ec, 

x + 9| x2 + lOx + 9 |x + l \ 



81x2+ 85X+4 
81x2 + 810x+729 
- 726ig 7 a 6 



X2 + 



>y 



/'te.F. 



9x + 9 

9x + 9 


Thus the H.C.F. is x + 1. 
This process may be performed in the following more compact form. 



2 


2x4 + 2x3-x2- 2x-l 


a^+ x^+ 4x + 4 


x2-9x + 81 




2x* + 2x3 + 8x + 8 
-x2 - lOx-9 


X4 + I0x3+ 9X2 




-1 


- 9x^- 9x2+ 4x 




x + 9 


x2 + 10x + 9 


- 9x3- 90x2- 81 X 






X2+ X 


81x2+ 85x + 4 






9x+9 
9x + 9 


81x2+810x +729 






-725x -725 


-725 







X +1 





Result: x + 1. 

2. x2 + 6 X - 7 and x^ - 39x + 70. 

3. x^ — X* — X + 1 and 5 x* — 4 x^ — 1. 

4. x8 + 2x2 + 9 and - 6x3 - iia;2 + i5x + 9. 

5. x3 - 2x2 - 15x + 36 and 3x2 - 4x - 15. 

6. x* - 3x3 + x2 + 3x - 2 and 4 x3 - 9x2 + 2 X + 3. 

7. 4x8 - 18x2 + 19x - 3 and 2x* - 12x3 + 19x2 _ 6x + 9. 

8. X* + 4x3 _ 22x2 - 4x + 21 and x* + 10x8 + 20x2 - lOx - 21. 

9. 6 a*x3 - 9 a3x2y - 10 a^ xy^ + 16 ay^ and 10 a^xV - 1^ «*«^y* + 8 a^*y^ 
12 a2x3y6. 



26 ALGEBRA TO QUADRATICS 

40. Least common multiple. The least common multiple of two 

or more polynomials is the polynomial of least degree that con- 
tains them as factors. We may find the least common multiple 
of several polynomials by the following 

EuLE. Multiply together all the factors of the various poly- 
nomials, giving to each factor the greatest exponent with which it 
appears in any of the polynomials, 

41. Second rule for finding the least common multiple. When 
only two polynomials are considered the previous rule is evidently 
equivalent to the following 

EuLE. Multiply the polynomials together and divide the 
product hy their highest common factor. 

EXERCISES 

Find the least common multiple of the following : 

1. x2 - y2,a;2 + ay — ax - xy, and a;2 - 2 xy + y\ 

Solution : x^ -y'^ = {x - y)ix -^ y). 

x'^ + ay -ax-xy = {x- y) {x — a). 
x^-2xy + y^ = {x- yf. 

Thus the L.C.M. = (x - yY (x + y) (x - a). 

2. 4a26c, 6a62, and 12 c2 

3. 9x?/2, 6x22/3, and ZxyH"^. 

4. (X + 1) (x2 - 1) and x3 - 1. 

5. x* + 4x22/2 and x2 + 2 2/2 - 2 y. 

6. 4x2 - 9?/ and 4x2 - 12x2/ + 9?/2. 

7. x2 - 4 X + 3, x2 - 1, and x2 - ax — x + a. 

8. x-1, 2x2-5x-3, and2x8-7x2 + 2x + 3. 

9. x* - 9x2 + 26x - 24 and x^ - 10x2 + 31 x - 30. 

10. 2x2 - 3x - 9, x2 - 6x + 9, and 3x2 _ 9x - 6x + 3&. 



CHAPTEE III 
FRACTIONS 

42. General principles. The symbolic statements of the rules 
for the addition, subtraction, multiplication, and division of alge- 
braic fractions are the same as the statements of the correspond- 
ing operations on numerical fractions given in (2), (3), and (4), 
§ 6. This is immediately evident if we keep in mind the fact 
that algebraic expressions are symbols for numbers and that if the 
letters are replaced by numbers, the algebraic fraction becomes a 
nunierical fraction. 

43. Principle I. Both numerator and denominator of a frac- 
tion may he multiplied (or divided) hy the same expression with- 
out changing the value of the fraction. 

This follows from (5), § 6. 

44. Principle II. If the signs of both numerator and denomi- 
nator of a fraction he changed, the sign of the fraction remains 
unchanged. 

This follows from Principle I, when we multiply both numerator and 
denominator by — 1 . 

45. Principle III. If the sign of either numerator or denomi- 
nator (hut not hoth) he changed, the sign of the fraction is changed. 

This follows from (6), § 6. 

46. Reduction. A fraction is said to be reduced to its lowest 
terms when its numerator and denominator have no common 
factor. We effect this reduction by the following 

Eule. Divide hoth numerator and denominator hy their 
highest common factor. 

27 



28 ALGEBRA TO QUADRATICS 

EXERCISES 
* Reduce the following to their lowest terms. 
12 ax2 - 12 ab^ 



1. 



4ax^-Sabx-\-4:al» 

12ax2-12a62 12 a {x - b) {x ■{■ b) 



4 ax2 - 8 a6x + 4 ab^ 





Solution : 



H.C.F. =4a{x-b). 

a;S + 4x *^ a — 2ax — lOx + 5 

g 6x2 -8ax +2a2 2 a26 + 2 a62 - 2 abc 

x2-a2 * ■ 3 6c2 - 3 62c - 3 abc ' 

g a2 + 62 _ c2 + 2 a& 21x8-9x2 + 7x - 3 

xi8 - ai8* • a2 - 62 4. c2 + 2 ac' ' 3x8 + 15x2 + x + 5 * 

^j^ 2x2+ 3x-9 j^2 X* - x8 - X + 1 j^„ x3 + 3ax2 + 3a2x + a8 
x2-9 ' ' 2x4 -x3-2x + l' * a2 + 2ax + x2 

47. Least common denominator of several fractions. We have 
the following 

Rule. Find the least common multiple of the various denomi- 
nators. 

Multiply both numerator and denominator of each fraction hy 
the expression which will make the new denominator the least 
common multiple of the denominators. 



EXERCISES 

Reduce the following to their least common denominator. 

- 2 8 ^ 2x-3 

1. -> , and — -. 

X 2x-l 4x2-1 

Solution : The L.C.M. of the denominators is x (4x2 — 1) . Thus the frac- 
tions are 

2 (4 x2 - 1) 3x(2x+l) ^^ x(2x-3) 

x(4x2-l)' x(4x2-l) ' ^^ x(4x2-l)' 
" , and 3. , ■ , and 



6 + a a2-62 2x-8 4x2 + 4x-16 4x2-26 



FRACTIONS 29 



A 1 1 ^ 1 

4. , , and 



aj3 _ y3 aj4 _ y4 a;2 - y2 



5. , — - — , and 



c 2x-\ X ^ 2x-3 

6. — , , and 



x2 - 2 X + 1 x2 - 1 (x + 1)2 

-a b . c 

7. : , , and 



a + b-c a-\-b -^c d^ + 2ab + b^ - c"^ 

48. Addition of fractions. This operation we perform as 
follows : 

EuLE. Reduce the fractions to he added to their least common 
denominator. 

Add the numerators for the numerator of the sum, and take 
the least common denominator for its denominator. 

49. Subtraction of fractions. This operation we perform as 
follows : 

Rule. Reduce the fractions to their least common denomi- 
nator. 

Subtract the numerator of the subtrahend from that of the 
minuend for the numerator of the result, and take the least 
common denominator for its denominator. 

50. Multiplication of fractions. This operation we perform 
as follows : 

Rule. Multiply the numerators together for the numerator 
of the product, and the denominators for its denominator. 

51. Division of fractions. This operation we perform as 
follows : 

Rule. Invert the terms of the divisor and multiply by the 
dividend. a 

Remark. Since a fraction is a means of indicating division, t~^^ and - 

are two expressions for the same thing. 

d 



30 ALGEBRA TO QUADRATICS 



EXERCISES ^ 

Perform the indicated operations and bring the results into their simplest 
forms. 

a + b a — b 



I I* — w a + b 

' a + b a — b 



a + & 



Solution ; 



a 4- 6 a — b 
a ^b a + 6 
a + b a — b 
a—b a+b 

a^-l^ 



{a + 6)2 + (a ■ 


-6)2 


a2-62 


(a + 6)2 _ (a . 


-6)2 


a2-62 




(a + 6)2 + (a - 


-6)2 


a2-62 


( 


2 a2 + 2 62 a2 + 62 


4a6 


2a6 


iW - 1 




2.V * 




3 + f + i 





(a + 6)2 - (a - 6)2 



' 1 + 1 ' 2.V * • 1-|-t\ 

5 2-V- g 3 + f + l 7 2 + f + f 

* i-(-W ■f + l + i.V '1-l + f" 

8. ^_^L±^. 9. ^-i^-^. 10. -i- + 



6 26 ab ac be a — b a + b 

11.1+1+1. 12. '?^l^+!-^ 13. 20= 



a6c c c a-la2-l 

-, 3a;-l 2«-7 -- 2ic-l 2x-5 

14. . 15. 

l-3x 7 a;-2a;-4 

16.—?^ L_. 17. ^ 8 



4x-4 6a; + 6 3x-9 6«-15 

18 « + ^ ■ g'^ + 2 a6 - 62 2a-36 3a-26 

' a-6 ■ a2-62 " " 12 a "*" 16a 

20 ?. ■ Q 21 «t? + ^c acZ-6c 



15(x-l) 10(x + l) 2cd{c-d) 2cd{c + d) 

22 a? + y a;-y 4zy ^^ a;2 + a;(a + 6) + a6 gg-gg 
' x-y x-\-y x2-y2' ' x2 - x(a + 6) + a6' x^ - 62' 

24. i5 + il + i?£-l^. 25. ^ « 



6 a 14 a 36 a 16 a a2 - 9 a + 14 ' a* - 6 a - 14 

26. Z^ + lU/^-l + lV 27 g(«-'g) a(a + x) 

\y8 a;/ \y2 y x) a2 + 2 ttx + x2 a2-2ax + xa* 

28. lzi^.lz^./i + _±_V 29. -A_ + _^ + _A._l. 
1 + 6 a + a« V 1-a/ (x - 1)« (x - 1)2 ^ x - 1 x 



FRACTIONS 31 



^ 1 

30. a + --L.. 31. -^^i 32. ^. 

c + ^ (^Y-1 x + l + l 



e \h/ X 

33. 1 34. 35. a + ^ 

1 .1 d 

x-\ 4h cH 

x-S 9 

a , h ^ db 1,1 a2 + 62 ^ 



o«a + 6 a-6 a2-62 i + x 1-x „„ a a2-62 

OO. • Of. • So. ' 

1 1 1111 a3 + 63 



(a + &)2 {a-hY 1 - « 1 + ic a h 

39 • a;-3y ^ x + 3y ^^ _J^ a; + 2 

■ x2-2a;y-15?/2 " x2-8xi/ + 15y2' ' 3(x + l) 3(-4-3x + x2) 

^^' 1 ri'+ 2y. ;• 

X y + z 
-Q 2a-36 + 4 3a-46 + 5 a-1 

6 8 12 

43 l ^"^ -^y^ x2 - y2 \ / x + y X - y \ 
■ \x2 - 2/2 a;2 4. 2/2/ • \^a; _ y x^yj' 

X- 1 y- 1 z -1 

-^ 3xy2 x__ y g 

' yz -{• zx — XV 11 1 

X y « 
45 / 2x + y 2y-x x2 \ x2 + y2 

\ X + y X - y x2 - y2/ ■ x2 - y2 
-^a — 36 4a — 6 5a + 3c a2 — 6c 2a 

40. 1 1 • 

6a 26 9c 2ac 6 

. „ 6cd! cda 

47. 1 ^ 

(a - 6) (a - c) (a - d) (6 - c) (6 -d^(b-a) 

dab ^ 



(c-6)(c-a)(c-d) 
a6c 
"^ {d -a){d- 6) (d - c) ' 



48 1 a-26 8 3a-4c2 9 5c2_--_66 
' 6a 3a6 46 8ac2 8c2 126c2 



' X-1 X + 1 (X - 1)2 (X + 1)2 X2 - 1 (X2 - 1)2 



CHAPTER IV 



EQUATIONS 

52. Introduction. An equation is a statement of equality 
between two expressions. 

We assume the following 

Axiom. If equals he added tOy subtracted frorriy multiplied hy^ 
or divided hy equals, the results are equal. 

As always, we exclude division by zero. In dividing an equation by an alge- 
braic expression one must always note for what values of the letters the divisor 
vanishes and exclude those values from the discussion. 

53. Identities and equations of condition. Equations are of 
two kinds : 

First. Equations that may be reduced to the equation 1 = 1 by 
performing the indicated operations are called identities. 

Thus 2 = 2, 

a -6= (3a -26) -(2a -6) 

are equations of this type. In identities the sign = is often replaced by =. It 
should be noted that identities are true whatever numerical values the letters 
may have. 

Second. Equations that cannot be reduced to the form 1 = 1, 
but which are true only when some of the letters have particular 
values, are called equations of condition or simply equations. 

Thus z=2 cannot further be simplified, and is true only when x has the value 
2. Also a; = 2 a is true only when x has the value 2 a or a has the value - • If in 
this equation x is replaced by 2 a, the equation of condition reduces to an identity. 

The number or expression which on being substituted for a 
letter in an equation reduces it to an identity is said to satisfy 
the equation. 

Thus the number 5 satisfies the equation x^ — 24 = 1. The number 3 satisfies 
the equation (x — 3) (a; + 4) = 0. 

32 



EQUATIONS 33 

The process of finding values that satisfy an equation is called 
solving the equation. The development of methods for the solu- 
tions of the various forms of equations is the most important 
question that algebra considers. 

In an equation in which there are two letters it may be possible 
to find a value which substituted for either will satisfy the equa- 
tion. Thus the equation x — 2 a = Ois satisfied if x is replaced by 

2 -a, or if a is replaced by -• In the former case • we have solved 

for £c, that is, have found a value that substituted for x satisfies 
the equation. In the latter case we have solved for a. In any 
equation it is necessary to know which letter we seek to replace 
by a value that will satisfy the equation, that is, with respect to 
which letter we shall solve the equation. 

The letter with respect to which we solve an equation is called 
the variable. 

Values which substituted for the variable satisfy the equation 
are called roots or solutions of the equation. 

When only one letter, i.e. the variable, occurs in an equation, the root is a num- 
ber. When letters other than the variable occur, the root is expressed in terms 
of those letters. 

54. Linear equations in one variable. An equation in which 
the variable occurs only to the first degree is called a linear equa- 
tion. To solve a linear equation in one variable we apply the 
following 

EuLE. Apply the axiom (§ 52) ^o obtain an equation in which 
the variable is alone on the left-hand side of the equation. 

The right-hand side is the desired solution. 

To test the accuracy of the work substitute the solution in the 
original equation and reduce to the identity 1=1. 

Since the result of adding two numbers is a definite number, and the same is 
true for the other operations used in finding the solution of a linear equation, it 
appears that every linear equation in one variable has one and only one root. 

When both sides of an equation have a common denominator, the numerators 
are equal to each other. This appears from multiplying both sides of the equation 
by the common denominator and then canceling it from both fractions. 



34 ALGEBRA TO QUADRATICS 

EXERCISES 

Solve : 

- 4«-2 , 6a; Sx ^ 

^•-^ + T = T + '- 

Solution : Transpose the term involving x, 

4a;-2 5x 3x ^ 

... . ^. 32x-16 + 25x-30a; ^ 

Add fractions, = 5. 

40 

Clear of fractions and simplify, 97 x — 216 

x = 8. 
rt a(d^ + x^) , ax 

dx d 

d^ 4- x^ X 

Solution : Divide by a, = c + -. 

dx d 

Transpose the term involving x, 

d* + x2 X 



dx d 



= c. 



Add fractions, = c. 

dx 

Clear of fractions and simplify, , _ 

d 

X = -' 

c 

3. {a-l)x = b-x. 4. {a-x){l-x) = z^-l. 

5. a(x-a2) = 6(x-62). 6. 2a; - fx = f « - 1 - |x + 2. 

7. 8x - 7 + X = 9x - 3 - 4x. 8. .617 x - .617 = 12.34 - 1.234x. 

9. 3(2x-.3) = .6 + 5(x-.l). 10. 7 - 5x + 10 + 8x - 7 + 3x = x. 
11. (x-3)(x-4) = (x-6)(x-2). 12. f {x\[|(|x+5)-10] + 3}-8 = 0. 
13. (H-6x)2+(2 + 8x)2 = (H-10x)2. 14. 6 = 3x+i(x + 3)-^(llx-37). 

15. 2(x + 5) (X + 2) = (2x + 7)(x + 3). 

16. (7ix - 2|) - [4| -UH- 5a;)] = 18^. 

17. 6x - 7(11 - X) + 11 = 4x- 3(20 -X). 

18. (a - 6) (X - c) + (a + 6) (« + c) = 2(&x + ad). 

19. 2x - 3(6 + f X) + ^(4 - X) - ^(3x - 16) = 0. 

20. 5x-2 = fx + fx + fx + T'ffa; + H« + i|a;. 

21. (a - 6) (a - c + x) + (a + &) (a + c - X) = 2 a*. 

22. 12.9x - 1.46X - 3.29 - .99x - llx + .32 = 0. 

23. 6.7x - 2^7.8 - 9.3x) = 5.38 - 4|(.28 + 3.6x). 



EQUATIONS S6 

24.3-^ = ^-. 25. -^+l^c. 

3 11 mx nx 



3 X 
26 



x-J:^x-^^ 27 iil^^lll^?. 

x-S x-^' ' f (6x + l) 3' 

28 t(a^-4) ^1 29 2x2-3x + 5 ^2 

'1(3x4-5) 6* ■7x2-4x-2 7* 

3o.» + l = 12 + l 31. i:i-%i = ^--l 

x2x9 J+x4^+x4 

„rt« + &aj_c + cZx ««25x — 2_52x — 5 

a + b ~ c + d' ' 3"7x-3~ 7 3x- 7* 

„- ox ex , /x , „- X + a & X - 6 a 

34. — + — -^ — = h. 35. — = + -■ . 

bag b a a b 

36. 5^ + i, = x-l. 3y 3x-19^5x-25^3 

a X — 13 X + 7 



3 12 1 3 2 



38. L4-L^ = -l--L. 39. ^—^ + ^-^i^ + '-:z^ = o. 



3 12 12 1 

2'^x 3'^x S'x"^ 



^^x-8x + 12 „, 18 ,-a(2x + l) Sox- 46 4 

4U. 1- = z H • 41. — = — • 

x + 2 x-8 x + 2 36 66 6 



.„ ax 6x 2a6 (a + 6)2x 

'4(6. i f- = < 

6 a a + 6 ab 

43. 8ix - - - 3|x- 4^x + 1 = 0. 

6 

.. 1 a + b 1 a — b 

44. + = + 

a + b x a — b x 

45 ^ (^ - ^) _L ^ + 8 _ 3(5x + 16) 
X- 7 X -4~ 5X-28 ' 

46.^ 



a 6 

^^^ox-A_^ 



X b^c — X «c2 — X _ 



5X-.4 1.3 -3x _ 1.8-8X 



2 1.2 

^48. -^^ + -1^- + -^ = 2. 
6x + 2 15x4-6 3x4-1 

49.I^^2-i(. + 3) + 6 = ^-(^±?>. 
3 6^ ' 2 



36 



ALGEBRA TO QUADRATICS 



50. 
51. 
52. 
53. 
54. 
55. 
56. 
57. 
58. 
59. 
60. 
61. 
62. 
63. 
64. 
65. 
66. 
67. 



5x-l 3x + 2 ic2-30x + 2 



3(x + l) 2(x-l) 6x2-6 

3x-2 . 7x-3 . x + 100 



= 10. 



x + 3 x + 2 x2 + 5x + 6 
36(x-a) x-62 6(4a + cx) ^Q 

5a 166 6a 

16X-27 x + 3 _ 6 + 3x 4x-7 

21 6 ~ 2 3 " 

a{b-x) b{c-x) _ a + 6 , /& , a\ 

6x ex X \c &/ 

5x-6 9-lOx 3x-4 3-4x 



10 

4-2x 

3 

3- 



6 
1.5x 



7 
4x2 



6x-3 X-.5 3(2x-l) 



^ + 



6(2x-5) 2(2x-6) 3(2x-6) 

x« + i-3x»-i 3x«-i-x« x« „ 

4x 4 2 

ax — be 6x — ac_cx — 62 x — a x 

ab c^ be c a 



3x 



X — 2a X — 26 X — 2c 



6 + c 



a+c— 6 a+6— c a+6+c 



2x« + 7x»-i 7x"-44x«-i 4X" + 27x'»-i 



9 5X-14 

3x + 3 /x + 1 



18 



a(x-3) 6(x-3) a2(x-l) 62(x-l) ^^ 
6 a 62 "^ a2 ~ • 



(m + n)2x nx _ c 3nx 

mb 6 m (a — 6) 6 



m (a — 6) 6 

4(13x-.6) 3(1.2-x)_9x + .2 5 + 7x 

6 "^ 2 -~20~"^~~r~"^'^- 

(^^±^(x-a) + ^^i:iA'(x-6) = 2a(2a + 6-x). 
6 a 

ax — 6 ex — d (6n + dm)x + (6p + dg) _ a e 
mx—p nx — q (mx - p) (nx - g) ~" m n' 



EQUATIONS 37 

55. Solution of problems. The essential step in solving a 
problem by algebra is the expression of the conditions of the 
problem by algebraic symbols. This is, in fact, nothing else than 
a translation of the problem from the English language into the 
language of algebra. The translation should be made as close as 
possible, clause by clause in most cases. In general the result 
sought should be represented by the variable, which for that 
reason is often called the unknown quantity. 

Example. What number is it whose third part exceeds its fourth part by 
sixteen ? 

Solution : " What number is it " is translated by x. Thus we let x represent 
the number sought. " Whose third part " is translated by - . " Exceeds its 

XX 

fourth part " is translated by , i.e. the third part less the fourth part 

leaves something. " By sixteen " gives us the amount of the remainder. Thus 
the translation of the problem into algebraic language is 
Let X represent the number sought. 

^-^ = 16. 
3 4 

This equation should be solved and checked by the methods already given. 



PROBLEMS 

1. What number is it whose third and fifth parts together make 88 ? 

2. What number increased by 3 times itself and 5 times itself gives 99 ? 

3. What is the number whose third, fourth, sixth, and eighth parts 
together are 3 less than the number itself? 

4. What number is it whose double is 7 more than its fourth part ? 

5. In 10 years a young man will be 3| times as old as his brother is 
now. The brother is 7| years old. How old is the young man ? 

6. A father who is 53 years old is 3 years more than 12 1 times as old as 
his son. How old is the son ? 

7. If you can tell how many apples I have in my basket, you may have 
4 more than ^, or, what is the same thing, 4 less than i of them. How many 
have I? 

8. If Mr. A received ^ more salary than at present, he would receive 
$2100. How much does he receive? 

9. A boy spends ^ of his money in one store and \ of what remains in 
another, and has 24 cents left. How much had he ? 



38 ALGEBRA TO QUADRATICS 

10. A man who is 3 months past his fifty-fifth birthday is 4^ times as old 
as his son. How old is the son ? 

11. In a school are four classes. In the first is ^ of all the pupils ; in the 
second, ^ ; in the third, j- ; in the fourth, 37. How many pupils are in the 
school ? 

12. A merchant sold to successive customers ^, |, and ^ of the original 
length of a piece of cloth. He had left 2 yards less than half. How long 
was the piece ? 

13. How may one divide 77 into two parts of which one is 2| times as 
great as the other ? 

14. The sum of two numbers is 73 and their difference is 15. What are 
the numbers ? 

15. A father is 4i times as old as his son. Father and son together 
are 27 years younger than the grandfather, who is 71 years old. How old are 
father and son ? 

16. The sum of two numbers is 999. If one divides the first by 9 and 
the second by 6, the sum of these quotients is 138. What are the numbers ? 

17. The first of two numbers whose sum is a is b times the second. 
What are the numbers ? 

18. If the city of A had 14,400 more inhabitants, it would have 3 times 
as many as the city of B. Both A and B have together 12,800 more than 
the city of C, where there are 172,800 inhabitants. How many are in 
A and B ? 

19. Two men who are 26 miles apart walk toward each other at the 
rates of 3| and 4 miles an hour respectively. After how long do they meet ? 

20. A courier leaves a town riding at the rate of 6 miles an hour. Seven 
hours later a second courier follows him at the rate of 10 miles an hour. 
How soon is the first overtaken ? 

21. A can copy 14 sheets of manuscript a day. When he had been work- 
ing 6 days, B began, copying 18 sheets daily. How many sheets had each 
written when B had finished as many as A ? 

22. The pendulum of a clock swings 387 times in 5 minutes, while that 
of a second clock swings 341 times in 3 minutes. After how long will the 
second have swung 1632 times more than the first? 

23. The difference in the squares of two numbers is 221. Their sum is 
17. What are the numbers ? 

24. If a book had 236 more pages it would have as many over 400 pages 
as it now lacks of that number. How many pages has the book ? 

25. A man is now 63 years old and his son 21. When was the father 
19 times as old as his son ? 



EQUATIONS 39 

' 26. If 7 oranges cost as much less than 50 cents as 13 do more than 
50 cents, how much do they cost apiece? 

27.. The numerator of a fraction is 6 less than the denominator. Dimin- 
ish both numerator and denominator by 1 and the fraction equals |. Find 
the fraction. 

28. The sum of three numbers is 100. The first and second are respec- 
tively 9 and 7 greater than the third. What are the numbers ? 

29. Out of 19 people there were f as many children as women, and 1^ 
times as many men as women. How many were there of each ? 

30. A boy has twice as many brothers as sisters. His sister has 5 times 
as many brothers as sisters. How many sons and daughters were there ? 

ySl. A dealer has 5000 gallons of alcohol which is 85% pure. . He 
/wishes to add water so that it will be 75% pure. How much water must 
he add? i 

V 32. How much water must be added to 5 quarts of acid which is 10% full 
strength to make the mixture 8|% full strength ? 

.^33. A merchant estimated that his supply of coffee would last 12 weeks. 
He sold on the average 18 pounds a week more than he expected, and it 
lasted him 10 weeks. How much did he have ? 

34. At what time between 3 and 4 o'clock are the hands of a clock point- 
ing in the same direction ? 

35. At what time between 11 and 12 o'clock is the minute hand at right 
angles to the hour hand ? 

^^6. A merchant bought cloth for |2 a yard, which he was obliged to sell 
for $1.75 a yard. Since the piece contained 3 yards more than he expected, 
he lost only 2%. How many yards actually in the piece ? 

37. A man has three casks. If he fills the second out of the first, the 
latter is still f full. If he fills the third out of the second, the latter is still 
I full. The second and third together hold 100 quarts less than the first. 
How much does each hold ? 

38. A crew that can cover 4 miles in 20 minutes if the water is still, can 
row a mile downstream in f the time that it can row the mile upstream. 
How rapid is the stream ? 

39. A cask is emptied by three taps, the first of which could empty it 
in 20 minutes, the second in 30 minutes, the third in 35 minutes. How long 
is required for all three to empty the cask ? 

40. A can dig a trench in f the time that B can ; B can dig it in f the 
time that C can ; and A and C can dig it in 8 days. How long is required 
by all working together ? 



40 ALGEBRA TO QUADRATICS 

56. Linear equations in two variables. A simple equation in 
one variable has one and only one solution, as we have abeady 
seen (p. 33). On the other hand, an equation of the first degree 
in two variables has many solutions. 

For example, Sx + 7y = l 

is satisfied by innumerable pairs of numbers which may be sub- 
stituted for X and y. For, transposing the term in y, we get 

X- 3 , 

from which it appears that when y has any particular numerical 
value the equation becomes a linear equation in x alone, and 
hence has a solution. Thus, when y = 1, ic = — 2, and this pair 
of values is a solution of the equation. Similarly, cc = — 9, y = 4 
also satisfy the equation. 

57. Solution of a pair of equations. If in solving the equation 
just considered, the values of x and y that one may use are no 
longer unrestricted in range, but must also satisfy a second linear 
equation, we get usually only a single pair of solutions. Thus 
if we seek a solution, that is, a pair of values of x and y satisfying 

Sx-\-7y = l, 

such that also 

x + y = -ly 

we find that the pair of values x = — 2, y = 1 satisfy both equa- 
tions. Any other solution of the first equation, as, for instance, 
x=—9, y = 4, does not obey the condition imposed by the second. 
Two equations which are not reducible to the same form are 
called independent. 

Thus 6a; -82/ -4 = 

and 3 a; — 4 y = 2 

are not independent, since the first is readily reduced to the second by transposing 

and dividing by 2. They are, in fact, essentially the same equation. On the other 

hand, . „ 

' X — 4y=2 

and 3 X — 4 y = 2 

are not reducible to the same form and are independent. Since dependent equa- 
tions are identical except for the arrangement of terms and some constant factor, 
all their solutions are common to each other. 



= -% 



EQUATIONS 41 

This principle we may state as follows : 

Two equations , i, , r. 

ax + oy + c = 

and a'% + &'?/ + C = 

are dependent when and only when 

£_ &^ _ c 

Independent equations in more than one variable which have 
a common solution are called simultaneous equations. 

Two pairs of simultaneous equations which are satisfied by the 
same pair (or pairs) of values of x and y and only these are called 
equivalent. 

Thus r3.+7a==l. ^^ ra==- 

are equivalent pairs of equations. 

58. Independent equations. We now prove the following 

Theorem. If A=0 and B = represent two in 
equations^ then the pairs of equations 

^ = ^'(1) and {'''■^f^'!: (2) 

are equivalent where a, b, c, and d are any numbers such that 
ad — be is not equal to zero. 

The letters A and B symbolize linear expressions in x and y. 
Evidently any pair of values of x and y that makes both A = and 
B = 0, i.e. satisfies (1), also makes aA-{-bB=0 and cA + dB = 0, 
i.e. also satisfies (2). We must also show that any values of x 
and y that satisfy (2) also satisfy (1). 

For a certain pair of values of x and y let 

aA-{-hB = 0, (3) 

cA + dB= 0. (4) 

Multiply (3) by c and (4) by a (§ 52). 

Then acA + bcB = 0, (5) 

acA + adB = 0. (6) 



42 ALGEBRA TO QUADRATICS 

Subtract (5) from (6) (§ 52), 

(ad-bc)B = 0. 
Thus, by § 5, either ad — be = ot B = 0. 

But ad — be is not zero, by hypothesis ; consequently -6 = 0. 
Similarly we could show that ^ = 0. 

Thus if we seek the solution of a pair of equations ^ = 0, 
jB = 0, we may obtain by use of this theorem a pair of equiva- 
lent equations whose solution is evident, and find immediately 
the solution of the original equations. 

59. Solution of a pair of simultaneous linear equations. The 

foregoing theorem affords the following 

KuLE. Multijply each of the equations by some number such 
that the coejfficunts of one of the variables in the resulting pair 
of equations are identical. 

Subtract one equation from the other and solve the resulting 
simple equation in one variable. 

Find the value of the other variable by substituting the value 
just found in one of the original equations. 

Check the result by substituting the values found for both 
variables in the other equation. 

Example. 

Solve 3x-l-7y = l, (1) 

x + y=-\. (2) 

Solution : Multiply (1) by 1 and (2) by 3, 

3x + 7?/ = l, 
3x + 32/ = -3 



Subtract, 








4y = 


4 


Substitute 


in (2), 






x + l = 


1. 
-1. 


Check: Substitute 


in (1), 




« = 


-2. 






3. 


(- 


- 2) + 7 . 1 = 


-6 + 7 



60. Incompatible equations. Equations in more than one vari- 
able that do not have any common solution are called incompatible. 



EQUATIONS 43 

Theokem. The equations 

ax-\-by = c, ^ (1) 

afx-\-b'i/ = c' '*(2) 
are incompatible when and only when aV — ba^ — 0. 

Apply the rule of § 59 to find the solution of these equations. 
Multiply (1) by a' and (2) by a. 

We obtain aa^x + a^by — ca\ 

aa'x -f- ab'y = ac'. 
Subtract, (ab' — a'b) y = ac' — ca'. 

If now ab' — a'b is not zero, we get a value of y ; but since under 
our hypothesis ab' — a'b = 0, we can get no value for y since divi- 
sion by zero is ruled out (§ 7). Thus no solution of (1) and (2) 
exists. 
Example. 

Solve Sx + 1y = l, (1) 

6x + 14y = l. (2) 

Solution : Multiply (1) by 2, 

Qx + Uy = 2 
6x+ Uy= 1 
Subtract, 0=1 

which is absurd. Thus no solution exists. 

61. Resum^. We observe that pairs of equations of the form 
ax -\- by -{- c = Oj 
a'x -f J'y + c' = 
fall into three classes : 

(a) Dependent equations, which have innumerable common 
solutions. ^ ^ ^ 

Tlien a'^b'^7'' W 

(b) Incompatible equations, which have no common solution. 
Then ^^, _ ^,j ^ ^^ ^^^ ^^^ -^ ^^^ ^^^^^ 

(c) Simultaneous equations, which have one and only one pair 
of solutions. 

Then ab' - a'b ^ 0. 



18 



44 ALGEBRA TO QUADRATICS 

EXERCISES 

Solve and check the following : 
. 2x + 6y = l, 2 4x-6y,= 8, , 6x + 8y 

' Qx + 1y = S. ' |x-2/ = f. •x + |y = 3. 

^ 7x-3y = 27, ^ 2x-|2/ = 4, ^ |y = ^x-l, 

* 6x-6y = 0. ■3x-|2/=:0. '^-^yrzfx-l, 

„ 5x-4y + l = 0, g 3x + 4y = 253, g 6x + 3y + 2 = 0, 

1.7x-2.22/ + 7.9 = 0. 'y = 5x. ■3x + 2y + l=0. 

j^Q x + my = a, j^j^ x + y = |(5a + 6), 

' x — ny = b. ' X — y = |(a + 6 6). 

-2 2x-3y = -5a, -, |x-i(y + l) = l, 

• 3x-22/ = -5&. • i(a; + l) + |(y-l) = 9. 

-- 3.5x + 2i2/=13 + 4fx-3.5y, -- 3x + 2 ?/ = 5a2 + a6 + 5&2^ 
' 2ix + .8y = 22^ + .7x-3i2/. " 3?/ + 2x = 6a2 - a6 + 662. 



16. 


3 8 

x+r 

16 _ 4 
X 2/ 


= 3, 
= 4. 




Hint. Ketain fractions. 


18. 


X - c 
y-C 


a 
6' 


r 




x-y = 


a - 


h. 


n/\ 


x + 2y 
2x-2/ 


+ 1 
+ 1 


= 2, 



17. 



1 1_6 

- 4- - — -1 
X y 6 

111 

X y Q 



3x + l_4 



19. 4-2y 3 
X + y = 1. 

- + ^ = c, 

20. r"^"^; 21. "^ ^ 

^^-^ + ^ = 5. ■ ^ + ^ = ci. 

X — 2/ + 3 «! 6i 

5 7 



22. 



24. 



x + 2y 2x + 2/ 

7 ^ 6 
3x-2~6-2/* 
.9x-.7y + 7.3 _ 
13X-152/ + 17" 
1.2x-.2y + 8.9 ^ ^ 
13x-16y + 17 
X y 1 



.2. 



a + 6 a — 6 a — 6 

X _ y _ 1 
a + 6 a — 6 a + 6 





x+1 a+6+c 


2.3 


y + l~a-6 + c' 




X — 1 a + 6 — c 




y-1 a-6— c 


25. 






y _6 


27. 


a- a - c 
a2 - X a^-y _ 



EQUATIONS 46 

^^^ y = 4 - 3x + x2. * • (4a; - 7) (X - 3) = y. 

„Q4Vx-3Vy = 6, g-xVa-y\/6 = a + 6, 

3 Vx — 4 Vy = 1. , X -\-y = 2 Va. 

22 xV2 + ?/V3 = 3V3, 32 4V^T7-5Vir^ = 7, 

'xV3-y\/2 = 2V2. "3 VxTT - 7 Vy^ = 2. 

«^ Vx Vy „> Vx - 3 Vy + 3 

12, ^^4 9 , 

—= + —= = 1. , + , = 4. 

Vx Vy Vx - 3 V 2/ + 3 

. ,, , a + 6 + 1 x + ly + 2 2(x-y) 

(a_&)[x + (a+&)y]=a-6+l. 3(x - 3)-4(y- 3) = 12(2y -x> 

62. Solutions of problems involving two unknowns. Tlie same 
principle of translation of the problem into algebraic symbols 
should be followed here as in the solution of problems leading to 
simple equations (p. 37). 

PROBLEMS 

1. The difference between two numbers is 3|. Their sum is 9|. What 
are the numbers? 

2. What are the numbers whose sum is a and whose difference is 6 ? 

3. A man bought a pig and a cow for $100. If he had given $10 more for 
the pig and $20 less for the cow, they would have cost him equal amounts. 
What did he pay for each ? 

4. Two baskets contain apples. There are 51 more in the first basket than 
in the second. But if there were 3 times as many in the first and 7 times as 
many in the second, there would be only 6 more in the first than in the 
second. How many apples are there in each basket? 

5. A says to B, "Give me $49 and we shall then have equal amounts." 
B replied, "If you give me $49, I shall have 3 times as much as you. 
How much had each? 

6. A man had a silver and a gold watch and two chains, the value of the 
chains being $9 and $25. The gold watch and the better chain are together 
twice and a half as valuable as the silver watch and cheaper chain. The 
gold watch and cheaper chain are worth $2 more than the silver watch and 
the better chain. What is the value of each watch ? 



46 ALGEBRA TO QUADRATICS 

7. What fraction is changed into ^ when both numerator and denomi- 
nator are diminished by 7, and into its reciprocal when the numerator is 
increased by 12 and the. denominator decreased by 12 ? 

8. A man bought 2 carriage horses and 5 work horses, paying in all 
$1200. If he had paid $5 more for each work horse, a carriage horse 
would have been only J more expensive than a work horse. How much 
did each cost? 

9. A man's money at interest yields him $540 yearly. If he had received 
\% more interest, he would have had $60 more income. How much money 
has he at interest ? 

10. A man has two sums of money at interest, one at 4%, the other at 5%. 
Together they yield $750. If both yielded 1% more interest, he would have 
$165 more income. How large are the sums of money ? 

11. A man has two sums of money at interest, the first at 4%, the second 
at 3^%. The first yields as much in 21 months as the second does in 18 
months. If he should receive \% less from the first and \% more from the 
second, he would receive yearly $7 more interest from both sums. What 
are the sums at interest? 

12. What values have a mark and a ruble in our money if 38 rubles are 
worth 14 cents less than 75 marks, and if a dollar and a ruble together 
make %\ marks ? 

13. A chemist has two kinds of acid. He finds that 23 parts of one kind 
mixed with 47 parts of the other give an acid of 84 1% strength and that 
43 parts of the first with 17 parts of the second give an 80f% pure mixture. 
What per cent pure are the two acids ? 

14. Two cities are 30 miles apart. If A leaves one city 2 hours earlier 
than B leaves the other, they meet 1\ hours after B starts. Had B started 
2 hours earlier, they would have met 3 hours after he started. How many 
miles per hour do they walk ? 

15. The crown of Hiero of Syracuse, which was part gold and part silver, 
weighed 20 pounds, and lost 1^ pounds when weighed in water. How much 
gold and how much silver did it contain if 19;^ pounds of gold and 10^ 
pounds of silver each lose one pound in water? 

16. Two numbers which are written with the same two digits differ by 
36. If we add to the lesser the sum of its tens digit and 4 times its units 
digit, we obtain 100. What are the numbers ? 

17. A company of 14 persons, men and women, spend $48. If each man 
spends $4 and each woman $3, how many men and how many women are in 
the company? 



EQUATIONS 47 

63. Solution of linear equations in several variables. This 
process is performed as follows : 

KuLE. Eliminate one variable from the equations taken in 
pairs, thus giving a system of one less equation than at first 
in one less variable. 

Continue the process until the value of one variable is found. 

The remaining variables may be found by substitution. 

Special cases occur, as in the case of two variables, where an infinite number 
of solutions or no solutions exist. Where no solution exists one is led to a self- 
contradictory equation on application of the rule. See exercise 17, p. 48. 

EXERCISES 

Solve and check the following : 

« + 2/ + 2 = 9, 
1. a; + 2y + 42 = 16, 
x + 3y + 9z = 23. 

Solution : x + y -[■ z = Q x -{■ y ■}- z— ^ 

x + 2y -{-iz = 15 x + 3y-f 9z=:23 

2/ + 32= 6 2y + 8z = U 

y + 4z= 7 

y + Sz = 6 

y -\-4z = 7 

2=1 

y + 3 = 6. 

y = 3. 

a; + 3 + 1 = 9. 

x = S. 

Check : 5 + 9 + 9 = 23. 

X + y = 37, x + y = xy, 

2.x + 2 =25, 3. 2x + 2z=x2;, 

y + z = 22. Sz + Sy = zy. 

Hint. Divide the equations by xy, xz, yz respectively. 

X + y 4- z = 17, X + y + z = 36, 

4. X + z - y = 13, 5. 4x = 3y, 

x + 2;-2y = 7. 2x = 3z. 

L3x-1.9y=.l, 2x+2y + « = a, 

6. 1.7y-l.lz = .2, 7. 2y-i-2z+x = 6, 

2.9x-2.1z = .3. 2z+2x + y = c. 



48 ALGEBRA TO QUADRATICS 





x + 2y = 6, 


8. 


y + 22; =8, 


z + 2u = ll, 




u + 2x = Q. 




y z 


10. 


i + l = 25, 

X z 




1 + 1 = 20. 
X y 




"y _oo 




iy-Zx--''' 


12. 


. ^' -15, 



9. 



X + y = m, 
y + z = a, 
z + u = n, 
u — x = b. 



11. 



xy _1 

; + y~5' 
xz 1 



x + z 6 

yz _ 1 

y-{-z~ 1 

y + 1 

13.?^ = 4, 
X-32 ' z + 1 

yz ^^2. 2; + 3_l 



4y-5z x + 1 2 

14. 312/ = X + 2 + 12, 15. X + 2 = 2| 2/ - 14, 



4^ 2 = X + y + 16. y + 2; = 3f X - 32. 

x + 2y-z=4.6, x + 2y + 32 = 15, 

y + 22;-x = 10.1, 17. 3x + 5y + 7z = 37, 
2 + 2x-y = 6.7. 5x + By + llz = 59. 

7x + 6y + 72 = 100, (x+2)(2y + l) = (2x+7)y, 

x_2y + 2 = 0, 19. (x-2)(3z + l) = (x+3)(32-] 
3x + y-2« = 0. (l+l)(«+2) = (y+3)(z+l). 



CHAPTEE V 
RATIO AND PROPORTION 

64. Ratio. The ratio of one of two numbers to the other is the 
result of dividing one of them by the other. 

a 
The ratio of a to 6 is denoted by a : 6 or by - • 

The dividend in this implied division is called the antecedent, 
the divisor is called the consequent. 

65. Proportion. Four numbers, a, b, c, d, are in proportion when 
the ratio of the first pair equals the ratio of the second pair. 

This is denoted by a : 6 = c : d or by - = - • 

The letters a and d are called the extremes, b and c the means, 
of the proportion. 

66. Theorems concerning proportion. If a, b, c, d are in pro- 
portion, that is, if 

a:b = G:doT- = -j (I) 

b d ^ ^ 

then ad = be, (II) 

b:a = d:c, ' (III) 

a:c = b:d, (IV) 

a -\- b : a = c -{- d : Cf (V) 

a — b : a = c — d : Cj (VI) 

a-{-b:a — b=:c-{-d:c — d: C^II) 

Equation (III) is said to be derived from (I) by inversion. 
Equation (IV) is said to be derived from (I) by alternation. 
Equation (V) is said to be derived from (I) by composition. 
Equation (VI) is said to be derived from (I) by division. 
Equation (VII) is said to be derived from (I) by composition 
and division. 

49 



50 ALGEBRA TO QUADRATICS 

67. Theorem. If a numher of ratios are equal, the sum of any 
number of antecedents is to any antecedent as the sum of the 
corresponding consequents is to the corresponding consequent. 

Let a\h — c\d = e\f= g '.h^ 



or 


a c e g 
'h~d~f~h' 


To prove 


a + c + e b-hdi-f 
g h 


If 


a c e g 
1~'d~f~h~''' 


we have 


a = hr, 




c = dr, 



g = hr. 
Divide the sum of the first three equations by the last and we get 
gj^c + e ^ b + d-\-f 
g h 

68. Mean proportion. The mean proportional between two num- 
bers a and c is the number b, such that 

a:b =^b : c. 
By (II), § 66, we see that ac = b^. 



EXERCISES 

If a : 6 = c : d, prove that : 

a2 . , c2 



1. a + h: = c + d: 

a + b 

Solution : By (V), § 66, 
Squaring, we get 



a + b c + d 

a -hb _c + d 
a c 

(g + b)^ ^ (c + d)^ 
a2 c2 ' 



a + b _ c + d 
^^ a2 ~ c2 ' 



a + 6 c + d 
a + b: - = c + d 



a + b c-\- d 



I 



RATIO AND PROPORTION 61 

2. a2 : 62 = c2 : d2. 3. a + b:c + d = a:c. 

4. ma:mh = nc: nd. 5. a2 : c2 = a2 + 62 . c2 + d^. 

6. a2 + 62 : _^ = c2 + d52 : _^. 7. Va2 + c^ : VPTd^ = a : 6. 
a + 6 c + cZ 

8. ma + n6 : ra + s6 = mc + n(i : re + sd. 

9. a + h + c + d:a — b-\-c — d = a + b — c — d:a — h — c + d. 

10. Find the mean proportional between a^ + c^ and 62 + ^2. 

11. Find the mean proportional between a2 + 62 + c^ and 62 + c2 + d\ 

Solve the following f or x : 

12. 20:96 = x: 57. 
14. x — ax:Vx = Vx : x. 

- „ Vx + 7 + Vx 4 + Vx 

10. — — ^;;^ = 

V^rp^ -Vx4-v^ a-6a6 ac 

Hint. Use composition and division. 

18. (?^^^ + „i)..^±^-„i = (a + bf:z. 
\a — I a + 6 



13. 


8 a6 : X : 


= 6c : 1| ac. 






15. 


\--/x 


l-3Vx = 


= 1 


4. 


17. 


a + & fl 


^2-62 
= X : 


a - 


-6 



CHAPTER VI 

IRRATIONAL NUMBERS AND RADICALS 

69. Existence of irrational numbers. We have seen that in 
order to solve any linear equation or set of linear equations with 
rational coefficients we need to make use only of the operations 
of addition, subtraction, multiplication, and division. When, 
however, we attempt to solve the equation of the second degree, 
x^ = 2, we find that there is no rational number that satisfies it. 

Assumption. A factor of one memher of an identity between 
integers is also a factor of the other memher. 

Thus let 2 . a = 6, where a and b are integers. Then since 2 is a factor of the 
left-hand member, it must also be contained in 6. 



Theorem. iVb rational number satisfies the equation x^ = ^. 
Suppose the rational number - be a fraction reduced to i' 
lowest terms which satisfies the equation. Then 




or a2 = 2 b\ (1) 

Thus, by the assumption, 2 is contained in a^j and hence in a. 
Suppose a = 2 a'. 

Then by (1) 4 a'^ = 2 b% 

or 2 a'^ = h\ 

that is, 2 must also be contained in ft, which contradicts the 

hypothesis that t is a fraction reduced to its lowest terms. 

The fact that the equation x^ = 2 has no rational solution is 
analogous to the geometrical fact that the hypotenuse of an 
isosceles right triangle is incommensurable with a leg. 

62 



IRRATIONAL NUMBERS AND RADICALS 63 

70. The practical necessity for irrational numbers. For tlie 
practical purposes of the draughtsman, the surveyor, or the 
machinist, the introduction of this irrational number is superflu- 
ous, as no measuring rule can be made exact enough to distin- 
guish between a length represented by a rational number and one 
that cannot be so represented. As the draughtsman does not use 
a mathematically perfect triangle, but one of rubber or wood, it 
is impossible to see in the fact of geometrical incommensurability 
just noted a practical demand from everyday life for the intro- 
duction of the irrational number. In fact the irrational number 
is a mathematical necessity, not a necessity for the laboratory or 
draughting room, as are the fraction and the negative number. 
We need irrational numbers because we cannot solve all quad- 
ratic equations without them, and the practical utility of those 
nimibers comes only through the immense gain in mathematical 
power which they bring. 

71. Extraction of square root of polynomials. This process, 
from which a method of extracting the square root of numbers is 
immediately deduced, may be performed as follows : 

EuLE. Arrange the terms of the polynomial according to the 
powers of some letter. 

Extract the square root of the first term, write the result as 
the first term of the root, and subtract its square from the given 
polynomial. 

Divide the first term of the remainder hy twice the root 
already found, and add this quotient to the root and also to the 
trial divisor, thus forming the complete divisor. 

Multiply the complete divisor hy the last term of the root and 
subtract the product from the last remainder. 

If terms of the given polynomial still remain, find the next 
term of the root hy dividing the first term of the remainder hy 
twice the first term of the root, form the complete divisor, and 
proceed as before until the desired number of terms of the root 
have been found. 



54 ALGEBRA TO QUADRATICS 

EXERCISES 

Extract the square root of the following : 

1. a* - 2 a^x + 3 aH^ - 2 ax^ + x\ 

Solution : a* - 2 a^x + 3 a'^x'^ - 2 ax^ + x* \a^ — ax + x^ 

a* 

2a2-ax| - 2 a^x + 3 aH^ - 2 ax^ + x* 
- 2 g^x + a^x^ 
2a^-2ax + x^\ 2 a^^ - 2 ox^ + x* 
2 g^x^ - 2 ax8 + x^ 

2. 1 + x. 3. 1-x. 

4. 3x2 _ 2x + x* - 2x3 + 1. 5. x* - 6x3 + 13x2 - 12x + 4. 

6. x* + 2/4 + 2x3?/ - 2x2/3 _ a;2?/2. 7. 9x* - 12x3 + 34x2 - 20x + 25. 

8. 49g4-42g36+37g262_i2a63+464. 9. 2g6- 2gc - 2&c H g2 + 62 4. c2. 

10. W*102 + v*U^ + U>4v2 + 2 W3u2|0 _|_ 2 t>8t«2w + 2 W?3m2u. 

72. Extraction of square root of numbers. We have the 
following 

Rule. Separate the mcmher into periods of two figures each, 
heginning o.t the decimal point Find the greatest number 
whose sqyiare is contained in the left-hand period. This is the 
first figure of the required root. 

Subtract its square from the first period, and to the remainder 
annex the next period of the number. 

Divide this remainder, omitting the right-hand digit, by twice 
the root already found, and annex the quotient to both root and 
divisor, thus forming the complete divisor. 

Multiply the complete divisor by the last digit of the root, 
subtract the result from the dividend, and annex to the remainder 
the next period for a new dividend. 

Double the whole root now found for a new divisor and pro- 
ceed as before until the desired number of digits in the root 
have been found. 

In applying this rule it often happens that the product of the complete divisor 
and the last digit of the root is larger than the dividend. In such a case we must 
diminish the last figure of the root by unity until we obtain a product which is 
not greater than the dividend. 



IRRATIONAL NUMBERS AND RADICALS 



55 



At any point in the process of extracting the square root of a number before 
the exact square root is found, the square of the result already obtained is less 
than the original number. If the last digit of the result be replaced by the next 
higher one, the square of this number is greater than the original number. 

There are always two values of the square root of any number. Thus Vi = + 2 
or — 2, since (+ 2)2 = (— 2)2= 4. The positive root of any positive number or 
expression is called the principal root. When no sign is written before the radical, 
the principal root is assumed. 

EXERCISES 

Extract the square root of the following : 
1. 2.0000. 



Solution : 


2'.00'00'00'|1.414 
1 
2.4|1.00 
96 










281 1 400 
281 






2.824 1 11900 


-A 




11296 
604 




2. 96481. 


3. 56169. 


4. 3. 


5. 877969. 


6. 2949.5761. 


7. 5. 


8. 257049. 


9. .00070128. 


10. 99. 


11. 69.8896. 


12. .0009979281. 


13. 12. 


14. 49533444. 


15. 9820.611801. 


16. 160. 



73. Approximation of irrational numbers. In the preceding 
process of extracting the square root of 2 we never can obtain a 
number whose square is exactly 2, for we have seen that such a 
number expressed as a rational (i.e. as .a decimal) fraction does not 
exist. But as we proceed we get a number whose square differs 
less and less from 2. 

Thus 1.2 = 1, less than 2 by 1. 

1.42 = 1.96, less than 2 by .04. 
1.41^ = 1.9881, less than 2 by .0119. 
1.4142 = 1.999396, less than 2 by .000604. 



5Q ALGEBRA TO QUADRATICS 

Though we cannot say that 1.414 is the square root of 2, we may 
say that 1.414 is the square root of 2 correct to three decimal 
places, meaning that 

(1.414)2 < 2 < (1.415)2. 

74. Sequences. The exact value of the square root of most 
numbers, as, for instance, 2, 3, 5, cannot be found exactly in deci- 
mal form and so are usually expressed symbolically. By means 
of the process of extracting square root, however, we can find a 
number whose square is as near the given number as we may desire. 
We may, in fact, assert that the succession or sequence of numbers 
obtained by the process of extracting the square root of a number 
defines the square root of that number. Thus the sequence of 
numbers (1, 1.4, 1.41, 1.414, • • •) defines the square root of 2. 

75. Operations on irrational numbers. Just as we defined the 
laws of operation on the fraction and negative nimibers (pp. 2-4), 
we should now define the meaning of the sum, difference, prod- 
uct, and quotient of the numbers defined by the seqilence of num- 
bers obtained by the square-root process. To define and explain 
completely the operations on irrational numbers is beyond the 
scope of this chapter. It turns out, however, that the number 
defined by a sequence is the limiting value of the rational num- 
bers that constitute that sequence, that is, it is a value from which 
every number in the sequence beyond a certain point differs by as 
little as we please. We may, however, make the following state- 
ment- regarding the multiplication of irrational nimibers : In 
the sequence defining the square root of 2, namely, (1, 1.4, 1.41, 
1.414, • • •) we saw that we could obtain a number very nearly 
equal to 2 by multiplying 1.414 by itself. In general, we multi- 
ply numbers defined by sequences by multiplying the elements of 
these sequences; the new sequence^ consisting of the products, defines 
the product of the original numbers. 

Thus (1, 1.4, 1.41, 1.414, • • ) (1, 1.4, 1.41, 1.414, • • •) 
= (1, 1.96, 1.9881, 1.999396, • • •). 

The numbers in this sequence approach 2 as a limit, and hence 
the sequence may be said to represent 2, 



IRRATIONAL NUMBERS AND RADICALS 57 

76. Notation. We denote the square root of a (where a repre- 
sents any number or expression) symbolically by Va, and assert 

or, more generally, ^ _ , 

Va • V^ = Va • b. 

Similarly, Va -^ V^ = Va -^ b. 

EXERCISES 

1. Form five elements of a sequence defining VS. 

2. Form five elements of a sequence defining V6. 

3. Form five elements of a sequence defining V6. 

4. Form, in accordance with the rule just given, four elements of the 
sequence \/2 • VS. Compare the result with the elements obtained in Ex. 3. 

5. Form similarly the first four elements of product V2 • V6 with the 
first four elements obtained by extracting the square root of 10. 

77. Other irrational numbers. The cube root and higher roots 
of numbers could also be found by processes analogous to the 
method employed in finding the square root, but as they are 
almost never used practically, they will not be included here. 
It should be kept in mind, however, that by these processes 
sequences of numbers may be derived that define the various 
roots of numbers precisely as the sequences derived in the pre- 
ceding paragraphs define the square root of numbers. 

The Tith root of any expression a is symbolized by Va. Here 
n is sometimes called the index of the radical. The principle for 
the multiplication and division of radicals with any integral 
index is given by the following 

Assumption. The product (or quotient) of the nth root of two 
numbers is equal to the nth root of the product (or quotient) of 
the members. 

Symbolically expressed, 

Va • -y/b = VoT^, 

Va H- V^ = Va -hb. 



68 ALGEBRA TO QUADRATICS 

78. Reduction of a radical to its simplest form. A radical is 
in its simplest form when the expression under the sign is integral 
(§ 11) and contains no factor raised to a power which equals 
the index of the radical ; in other words, when no factor can be 
removed from under the radical sign and still leave an integral 
expression. We may reduce a quadratic radical to its simplest 
form by the following 

Rule. If the expression under the radical sign is fractional^ 
multiply both numerator and denominator by some expression 
that will make the denominator a perfect square. 

Factor the expression under the radical into two factors^ one 
of which is the greatest square factor that it contains. 
■ Take the square root of the factor that is a perfect square, and 
express the multiplication of the result by the remaining factor 
under the radical sign. 

If the radical is of the nth index, the denominator must be made a perfect nth 
power, and any factor that is to be taken from under the radical sign must also be a 
perfect nth power. 

EXERCISES 



Reduce to simplest form : 




1. Vv. 

Solution : 


/T2 /12 . 5 /4 • 15 
\5=\ 25 =\ 25 ~- 


= |Vl6. 
5 


2. vi- 


3. V32. 


4. Vf. 


5. V27. 


6. VA. 


7. V243. 


8. V'250. 


9. Vi + 4. 


10. Vi-|. 


11. 8V75. 


12. tV2762. 
15. V¥ + |. 


13. lV80a;8y*. 


14. VtV + ^V 


16. v-V^ + ^V 


"■ xS- 


■ ^^ S 


■'■ xlf ■ 


"M 


-->/!• 


«■f^/S■• 


-# 


-• ^^m■ 


- -Vi^- 



IRRATIONAL NUMBERS AND RADICALS 69 



26. Vx^ - 2 x2y + X2/2. 27. V6x8-20x2 + 20x. 



28 M^-^«'^ + « 29 / 2a;«-I2x2:+l8x 

„Q / 2 gs - 8 a2 + 8 g „- j a^ + a% -~ab' 

' \8x-8x2 + 2x3' * \ 9(g-&.) 



g62 _ 63 



79. Addition and subtraction of radicals. Radicals that are of 
the same index and have the same expression under the radical 
sign are similar. Only' similar radicals can be united into one 
term by addition and subtraction. We add radical expressions by 
the following 

Rule. Reduce the radicals to he added to their simplest form. , 
Add the coefficients of similar radicals and prefix this sum 
as the coefficient of the corresponding radical in the result. 

A rule precisely similar is followed in subtracting radical 
expressions. 

EXERCISES 

Add the following : 

1. V27, \/48, and V75. 
Solution : 



V27 = - 


n/ 9 • 3 = 3 V3 


V48=- 


v/I6.3= 4V3 


\/75=^ 


v/25.3= 6V3 



Sum = 12 \/3 

2. V3 + 2V3. 3. 8V7-3V7. 4. ay/x-hy/x. 

5. g + 2Vg + 3\/a + 2 Vl6g - \/27g. 

6. 3 V8 + 4 V^ ~ 5 V50 - 7 V72 + 6 V98. 

7. 8Vg+5\^-7Vg + 4Vg-6Vx-3 Vg. 

8. 7 V4x + 4 Vox + 3 ViSx - 5 V36x - 2 V80x. 

9. Vg^^ + Vl6g - 106 + Vgx2 - 6x2 _ V9(g - 6). 



10. 4 Vo^ - 3 y/¥x + 2 Vc^ + Vd2x - 2 V(6 + dfx. 

11. 6 Vx + 3 V2x - 5 V3x - 2 V4x + Vl2x - \/l8x. 



60 ALGEBRA TO QUADRATICS 

80. Multiplication and division of radicals. For these pro- 
cesses we have the following 

EuLE. Follow the usual laivs of operation (§ 10), using also 
the assumption of § 77. 

Beduce each term of the result to its simplest form. 

" The operations of this section are limited to the case where 
the radicals are of the same index. Radicals of different indices 
as Vs and V^ must first be reduced to the same index. See § 87. 

EXERCISES 

1. Multiply V2 - V3 by V2 - VS. 
Solution : V2 - V3 

V2-V8 



2 _ V6 - Vl6 + V2i 
= 2-V6-4 + 2V6 

= - 2 + V6. 



2. Divide ^_tZ_ by v^ + \/y. 
Vxy 
x-\-y 

y/xy ( Vx2 - Vxy + Vy2) ( Vx + y/y) * 
Solution : — — —— = — ^ ^ 

Vx + y/y y/xy ( Vx + Vy) 

_ Vx2 Vxy Vy2 
Vx2/ Vxy V^ 



<^--</!- 



Carry out the indicated operations and simplify : 
3. VlO . V5. 4. V^ . ■^. 

5. V28 . V7. • 6. Vf ^ Vf- 

7. (a -6x^1 8. VS-Vff. 

9. ( V7 - V3) ( V3 - V2). 10. (- 1 + V3)^ 

11. (6 V3 + V6) (6 V2 - 2). 12. y/lTc • VTOc. 

* Since as + 63= (a + ft)(a« - a6 + ft*), if o= V^, 6= Vy we have 



IRRATIONAL NUMBERS AND RADICALS 61 

13. (a + 6-V^)(Va+V&). 14. (8 + 3 Vs) (2 - VB). 



15. VVx + Vy • V Vx - s/y, 16. V6 + 2 V6 . V 6 - 2 V5. 



17. Vx + Vjc2 - 1 . Vx - Vx2 - L 18. (x2 + y2) ^ (a; ^ _,. y ^), 



19. --V^. 20. (4Va-V3^)(V^ + 2V3x). 



X 

21. VS.Jp. 
\4a 



25. (^.^Ij. 36.V^.Vg 



27, ^*-^ 



+ 66 
y ' \x/ \ ox^ — to2 

:fl^.(.t^-,%). 28. (5-l^).(l - 1). 

Vxy \y ^1 VVx Vy/ 

29. (2 V6 - Vl2 - V2i + Vis) V2. 

30. (3\/8 + Vl8+V60-2\/72)V2. 

31. (5 v^ - 4 V32 + 3 V60 - 3 V54) V3. 

32. ( V9^T6 + 3 A^)(V9x + 6 - 3 \^). 

33. [(V7 4 V3 + Vl0)(V7 + V3-Vi0)7. 

34. (2 V30 .. 3 V6 + 5 V3) ( V8 + V3 - Vs). 

35. (2V^ + V8-Vl2)(^V30-fV3+V2). 



37. Find the value of ^ V2i - V| + 2 Vs - V6- V3 4- V5 to three 
decimal places. 

81. Rationalization. The process of rendering the irrational 
numerator (or denominator) of a fractional expression rational 
without altering the value of the fraction is called the rationaliza- 
tion of the numerator (or denominator) of the fraction. 

This is usually accomplished by multiplying both numerator 
and denominator of the given fraction by a properly chosen 
radical expression called the rationalizing factor. 



62 ALGEBRA TO QUADRATICS 

The principles in accordance with which this rationalizing 
factor is selected are the following: 

Principle I. Since {a ■i-b)(a — b)=a^ — b% the rationalizing 
factor of -sfx ± Vy is sjx hF Vy. 

Principle II. Since {a^ — a'b-\-lF) (a + 5) = a' + h^, the ration- 
alizing factor of -sfx -\- Vy is V^ — V^ + Vy^ and conversely. 

Since (a^ -\- ah -\- If) (a — b)=a^ — b^, the rationalizing factor 
of -y/x — -Vy is V^ + Vxy + Vy^, and conversely. 

EXERCISES 

^Nationalize the denominators of the following : 
- Va + Vx 
Va — Vx 
Solution : By Principle I the factor which will render the denominator 

rational is Va + Vx. 

Va + Vx Va + Vx Va + Vx a + x + 2 Vox 



Thus 
2. 



Va—Vx Va—Vx Va + Vx 
1 



2+V2 + V3 
Soltttioii : This problem requires a twofold rationalization. 

1 (2+V2)-V3 

2 + V2 + V3~ [(2+ V2) + V3][(2 +\^) - V3] 
^ 2+-v^-V3 ^ 2+V^-V3 
~ (2 + V2y-S ~4 + 4V2 + 2-3 
_ 2+V2-V3 _ (2 + v^-V3)(3-4\^ 

3 + 4v^ ~ (3 + 4V2)(3-4v^) 
_ 6 + 3-v^-3V3-8\/2-8 + 4V6 
"" 9 - 16 . 2 

2 + 6v^+3V3-4V6 



^2-V3 
Solution : 



^2-2 + ^6 + ^^ Vi + v'e + v^ 



^-.■^S ^-^3 ^2+^6+V3-« 2-3 

= ^(^ + V6 + ^9). 



IRRATIONAL NUMBERS AND RADICALS 63 

5 7-V5 g V3+V2 



2+V3 3+V6 V3-V2 



8 /- 3, 



V3+V2 V2-V4 V2+3Vi 

10. ^/^«±^. 11. I^±^^. 

\a-y/x \a-Va2-l 

12 2V6 ^2 1+3V2-2V3 

* V2+V3+V6 ' V6_}.V3+V2 

-- 2\/T6 -g Va + ic + Va — x 



V3 + V5 + 2 \/2 Va + x - Va - x 

16 2 j^ V6-V5-V3 + V2 

Va + 1 + Va- 1 V6 + V5 - V3 - V2 

18. Show that ^^ ~ "^ = - .10 • ■ -. 

V2-fV3 

19. Show that ^^^-^^ = 17.48 .... 





V8- 


-V7 






20 


V(l + a)(H-6)- 


-V(i 


-a)(l 


-&) 


21. 


V(l + a)(l + 6) + V(l 
Show that ^^+^^ - 


-a)(l 


-h) 

V6_ 



= .168. 



V3 + V5 + V6 - V5 

82. Solution of equations involving radicals. We prove the 
important 

Theorem. WTien an equation in x is multiplied hy an expres- 
sion in Xy the resulting equation has, in general, solutions which 
the first one did not possess. 

Let ^ = 

represent an equation contaming x which is satisfied by the 
values X = a, by ■ - n. Let 5 be an expression which vanishes 
when X = ay p, • ' V. Then the expression 

is satisfied not only when x = a, bj • - j n, but also when 
x = a,l3,-';v. . 



64 ALGEBRA TO QUADRATICS 

Example. The equation x — 2 = 

has X = 2 for its only solution, while the equation , 

(X - 2) (X - 3) = 
has in addition the solution x = 3. 

If in the course of a problem it is necessary to multiply an 
equation by any expression involving the variable, the solutions 
of the resulting equation must be substituted in the first one to 
ascertain if any solutions have been introduced which did not 
satisfy the original equation. Solutions which have been intro- 
duced in the process of solving an equation, but which do not 
satisfy the original equation, are called extraneous solutions. 

It may be shown in a similar way that raising the equation in Xj 

A =B, 
to any power introduces extraneous solutions. 

EXERCISES 



1. Solve Vaj + 19 + Vx + 10 = 9. 

Solution : '^ Vx + 19 = 9 - Vx + 10. 

X + 19 = 81 + X + 10- 18 V»Tl0. 
-72=- 18 Vx + 10. 
4 = Vx + 10. 
16 = x + 10. 
x = 6." 
Check : V6 + 19 + V6 + 10 =5 + 4 = 9. 

2. Solve Va; + 19 - Vx + 10 = - 9. 

Vx + 19 = - (9 - VxTlO). 
Simplifying, we get x = 6. 



Check : V6 + 19 - V6 + 10 =+6-4 = 1 9^-9. 

Thus our result satisfies only the equation which was introduced in the 
course of solving the problem, and is extraneous. The Original equation 
has no solution. 

Solve and check, noting all extraneous solutions : 

3. VSx -1=5. • 4. Vi« -8 = 2. 

5. V2x + V3x = 1. 6. Vx + V3x = 2. 

7. V5x-7=v'4x + 3. 8. 5Vx-7 = 3v^-l. 



IRRATIONAL NUMBERS AND RADICALS 



65 



9. Vx + 1 + Vx + 2 = 3. 
11. 2\^-V2x = 2+v^. 
13. V37 - 7 V6x^=^ = 4 



10. Vl3 4-4v^^^ 



6. 



12. Vx + 4 + \^+T=L 



14. Vx2 - 7 x + 19 = v^^^. 



15. 7 + Va;2-lla; + 4 = jc. 



17. a; — Vax (1 + x) + 1 - x = 1. 
19. |(7V5 + 6)-5=|(3v^-l). 
I 4 



16. 8 + V(x - 10) (X - 6) = X. 
18. V2 (X + 1) + V2x + 15 = 13. 
20. 2\/3 + 3V2x = 3V2 + 2V3a 



21. 



6 4- Vx 8 



23. V7 X + 2 = 



25. V2X-1 



-Vx 

6x + 6 
V7X4-2' 

2 (x - 3) 
V2x-10' 



22. 



24. 



4. 



5 + v^ 
5-Vx 
a — Vto 26 



3 Vox 



a + V6x 2 6 + 3 Vox 
6x+ 10 



26. V9 X + 10 = 



27. 



VTT- 



VT 



1 + VI -X 1 - VIT^ 

29. X — ax : Vx = Vx : X. 

30. 2v^^T2 + V^T2 = i^^±^. 

V8x + 8 

1 + 2V3X-6 11 + 2 V3x- 5 



V4X+9 



28. Va - X + V6 - X = 



V6 



31. 



1 + 3V3X-6 11 + 5V3X-5 



32. V9x + 7 + V4X + 1 = V25x + 14. 

33. Vx + 15 + Vx - 24 - Vx - 13 = Vx. 



34. Vx - 7 + Vx - 2 - Vx - 10 = VxT5. 

35. (VS-7)(Vx-3) = (Vx-6)(Vx-5). 

36. (a + \^) Vx:(6-Vx) Vx = a + 1:6-L 

37. (4V5-7):(5V^-6) = (V^-7):(V^-6). 

38. ( Va V6 — V6 Va) Vx = a V6 Vx — 6 VaVx. 



CHAPTEE VII 

THEORY OF INDICES 

83. Negative exponents. We have already seen (§ 16) that 

a^-a"* = «" + "' (1) 

when n and m are positive integers. We now assume that this law 
still holds when one or both of the numbers m and n are negative 
or fractional. 

If we let a-m = , 

a'" 



then 






since the law (1) holds when n and m are any integers. This 
notation may be expressed verbally as follows : 

Principle. A factor of numerator or denominator of a frac- 
tion may be changed from the numerator to the denominator, 
or vice versa, if the sign of its exponent he changed. 

84. Fractional exponents. Since (p. 57) Va • Va = a, it is 
natural to devise a notation for Va suggested by the law (1). 

If we let Va = a*, 

we have Va • Va = a* • a^ = a' "^ * = a^ = a. 

Furthermore, if we let -y^ — an 
it would be consistent with law (1) to write 

1 11 11 s 

(a")'' = a« . a» = a" " = «"'. 
This notation we shall assume in general. Thus 

66 



THEORY OF INDICES 67 

With the adoption of this notation we can attach a meaning 
:o any real number with any rational number for its exponent. 
This notation may be expressed verbally in the following 

Principle. The numerator of a fractional exponent indicates 
a power, the denominator a root. 

85. Further assumptions. The operation of multiplication is 
subject to the following laws of exponents : 

I. Commutative law of rational exponents : 

(^a'^y = a""- = a'--'' =(a'-y. 

II. Associative law of rational exponents : 

(yj = a'?'- -^ = ««"•« = (a'^y. 

The laws of operation (§ 10) defined for integral values of the 
symbols we also assume when the symbols are expressions with 
rational exponents. 

86. Theorem. a''lf= {ctby, where r is any rational number as _. 

We raise both sides of the equation a'b'' = (aby to the g-th 
power separately and show that the results are equal. 

Since r = -> 

l(abyY = [(abyj = {aby = aPbP. 

p p p p p p p p 

Also (or by = (a«55) = (a^^»«)(a«6«) • • • {a'^^) 







q terms 










p p 


P P P P 










q terms 


g terms 






=iJm' 


== a^hP, 








{ariry 


= l(abyj. 










ing the qih 


root and taking the principal 


root, 


we 


obtain 


a'b"- 


= (aby. 


• 









68 ALGEBRA TO QUADRATICS 

EXERCISES 
yl. Express in simplest form with positive exponents : 

, , 36a-26-ic-5 

(a) 

9a26-2c-i 

Solution : By Principle, § 83, ^^^"^^~^^"^ 
9a26-2c-i 

_46-i&+2c-ic+3 
~ a2-a2 

^46c 
~ a* ' 

(b) (c) — - 



^ ^ (xi2/l)-i ' 35x-2?/6z-4* 

3a-i6-2 6a2x-i gft^c Va6v^ 

4X-22/-4' 56-ic2' ^'^ v'^&-^aJ62c*' 

,.. 5 11 / — I — \ — ; /., V 4x-«y-3 15a'263-m 

^"•^ • ' 5 a- 4 6-"' 14x«2/«-3 

• 2. Arrange in order of magnitude the following : 

(a) Vi, 71, 7|. 

Solution : We first ask. Is (f )i > (|)^ ? 

Raise both numbers to the sixth power. 

We obtain (|)3 and (f)2, 

or If and f, 

er 2if and 2|. 

Thus Vl>7l- 

Now compare (f)^ and (|)io 

Raise both numbers to the fourth power. 
We obtain . (|)2 and |, 

or 1| and If. 

Thus VI > Vl. 



THEORY OF INDICES 69 

Now compare (§)' and (|)^. 

Raise both numbers to the twelfth power. 
We obtain (f)* and (|)3, 

or fl and \\^-, 

or 6yV and 6f|. 

Thus -s/i>\^h 

The order of magnitude is then Vli VI? Vf • 

(b) V^, VI, </h (c) 4^, V8, -Ws. 

(d) Vli, VI- (e) V3, ^5, Vl5. 

3. Perform the indicated operations, 
(a) \/2 . V3. 

Solution : ^2 • V3 = 2' • 3^ = 2^ 3^ 

= (22 . 38)i = V'4^ = v^lOS. 

V3 

V5 . ''V3.V2 

87. Operations with radical polynomials. These operations 
follow the rules for the same operations previously given, pro- 
vided the assumptions and principles of §§ 83-86 are observed. 

EXERCISES 

1. Divide x^ — y^ by ^/x — y/y. 

2. Extract the square root of 4 x - 12 x^ 2/3 + 9 2/5 + 32 x^ - 48 yi + 64. 
3x + 3x-i-6 



3. Simplify 



xi- 3x2 + 3x-^-x-i 



4. Divide J-Va^ -J-V^ by sJ-> 

a'2 52 

5. Extract the square root of 1 2. 

6. Multiply - 3x-5 + 2^ by ^ - ?^. 

X* x^ 6-1 

7. Extract the square root of 

-ix2y-2- i^yx-i + ^y2x-2-^xy-^ + 25?. 
49 2 Id 7 7 

8. Multiply V^ - x» + x^ f- Vx^ -x+v^-lby Vx + 1. 



QUADRATICS AND BEYOND 

CHAPTEE VIII 
QUADRATIC EQUATIONS 

88. Definition. An equation that contains the second bnt no 
higher power of the variable is called "a quadratic equation. The 
most general form of the quadratic equation in one variable is 

ax'^ + hx-{-c = 0, (1). 

where we shall always assume a, b, and c to represent rational 
numbers, and where a =^ 0. Every quadratic equation in x can 
be brought to this form by transposing and simplifying. 

89. Solution of quadratic equations. The solution of a quad- 
ratic equation consists in finding its roots, that is, the numbers 
(or expressions involving the coefficients in case the coefficients 
are literal) which satisfy the equation. 

The common method of solving a quadratic equation consists in 
bringing the member of the equation that involves the variable 
into the form of a perfect square, i.e. into the form 

x^-\-2Ax + A\ 

For example, let us solve 

Transpose 8, x^-^2x = S. 

If now we add 1 to both sides of the equation, the left-hand 
member will be a perfect square, 

x^-\-2x-{-l = 9. 
70 



QUADRATIC EQUATIONS 71 

Express as a square, (x + 1)^ = 9. 

Extract the square root, x -\-l =±S. 

Transpose, ic = — 4 or 2. 

Both — 4 and 2 satisfy the equation, as we see on substituting 

them for x. Thus 

(-4)^ + 2(-4)-8 = 0, 

and 2^ + 2 • 2 - 8 = 0. 

Consider now the general case. 

Let us solve ax'^ -{- bx -\- c = 0. 

Transpose c, ax^ -\- bx =— c. 

b c 
Divide by a, x^ -^ - x = 

/ bV 
Add ( 77— I to both members to make the left-hand member a 

\2aJ 

perfect square, 

b b'' _ c b"" _ -4.ac-\-b^ 

,a 4a^ a 4a^ 4 a^ 

4 ac 



/ by b^-4 

Express as a square, ( x + — — I = — —^ 
Extract the square root. 



b V^^ — 4 ac 



x-\-w- = ± 



2a 2a 

2a 



Transpose, x = ^^ * (1) 

The roots are 



-b-\- V^^ - 4 ac _ -b - -s/b^ -4.ac 
Xi — „ f X2 — „ 

2 a 2a 

That the equation can have no other roots appears from § 96. 

* This expression for the roots, _ / — 

X = » 

2a 

may be used as a formula for the solution of a quadratic equation. 

Thus to solve the equation 2a;2-3a; — 6=0 

we may substitute in the formula a = 2, 6 = - 3, c = — 6, and obtain 

3±\^+48 3±V57 



Thus Xi-. 



4 4 

3+V57 3-V57 



72 QUADRATICS AND BEYOND 

One should verify the fact that both a-iid 

— satisfy (1) and are consequently roots of the 



2a 

equation. They are, in geueral, distinct from each other. For 
particular values of the coefficients to be noted later (§ 98) the 
roots may be equal or complex (i.e. of form a -\- /3 V— 1, where 
a and yS are ordinary rational or irrational nimibers). 

We may sum up the process of solving a quadratic equation in 
the following 

EuLE. Write the expression in the form aa^ -\- hx -\- c = 0. 

Transpose the term not involving x to the right-hand side of 
the equation. 

Divide both sides of ^he equation by the coefficient of a^. 

Add to both members the square of one half of the coefficient 
of X, thus making the left-hand member a perfect square. 

Rewrite the equation, expressing the left-hand member as the 
square of a binomial and the right-hand member in its simplest 
form. 

Extract the square root of both members of the equation, not 
omitting the ± sign in the right-hand member. 

Transpose the constant term, leaving x alone on the left-hand 
side of the equation. The two values obtained on the right-hand 
side by taking the + and — signs separately are the roots sought. 

Check by substituting the solutions in the original equation, 
which should then reduce to an identity. 

90. Pure quadratics. A quadratic equation in which the coeffi- 
cient of the term in x is zero is often called a pure quadratic. Its 
solution is found precisely as in the general case, excepting that 
we do not need to complete the square. Thus let us solve 

ax^ -f- c = 0. 

Transpose c, ax^ =— c. 

Divide by a, ic* = — -. 



QUADRATIC EQUATIONS 73 



Extract the square root, ^ = ± _,' 
The roots are Xi = + -v/— -? x^ = ~ a/ 



EXERCISES 

Solve and check the following : 
1. 3x2 -6x- 10 = 0. 

Solution : Transpose 10, 3 x^ — 6 x = 10. 

. Divide by 3, x2 - 2 x = J/- 

Add the square of \ the coefficient of x, i.e. 1, to both sides, 

x2_2x + l = -Lo + l = Y- 
Express as a square, (x — 1)2 = -LS-. 

Extract the square root, x — 1 = ± "V^- 

x = i±V¥. 

Check: 3(l±^'-6(l±^)-10 = 0. 



3±6^| + |-6:F6Vir_io = 0. 



/13 



2. 8x2 + 2x -3 = 0. 

Solution : Transpose 3, 8 x2 + 2 x = 3. 

Divide by 8, x2 + i x = f . 



Add the square of \ the coefficient of x, i.e. ^^j, to both sides, 




^' + \^ + -h = \ + iz = il' 


Express as a square. 


(^ + \Y = f f- 


Extract the square root 


X + 1 = ± |. 




^=±'J-\ 




= -lov\. 


Check: 8. (1)2 + 2. 1- 


3 = 1 + 1-3 = 0. 


8(-f)^ + 2(-f)- 


-3 = 11-1-3 = 1-1-3=1-1 


3. x2-ax = 0. 


4. x2 = 169. 


5. x2-ix = i 


6. fx2 = 560. 


7. x2 + x-l = 0. 


8. 19x2 = 5491. 


9. 3x2-7x = 16. 


10. x2 = .074529. 


11. 3x2+ ll = 5x. 


12. x2 - tl X = 1. 


13. x2 + X - 66 = 0. 


14. 20x2 + x = 12. 



1 = 0. 



74 QUADRATICS AND BEYOND 

15. 7a;2 + 9x = 100. 16. 6x2 + 6x = 66. 

17. 14x2 - 33 = 71x. 18. 5x2 + 13 = 14x. 

19. x2 - 8x + 15 = 0. 20. 91 x2 - 2 X = 45. 

21. x2 + 2 X - 63 = 0. 22. x2 - 6x + 16 = 0. 

23. x2 - lOx + 32 = 0. 24. 6x2 + 26^ = 25 ix. 

25. 6x2 - 13x + 6 = 0. 26. 15x2 + 527 = 178x. 

27. 2x2 + 15.9 = 13.6 X. 28. (x - 1)2 = a(x2- 1). 

29. a'^{b-x)^=h^{a-x)^. 30. 13x2-19 = 7x2 + 5. 

31. ax2 - (a2 + l)x + a = 0. 32. (a - x)(x - 6) = - ab. 

33. 14 x2 + 45. 5 X = - 36. 26. 34. a^ {a - x)2 = b'^{b- x)2. 

35. (a-x)2+(x-6)2 = a2 + 62. 35. (a -x)(x-6) = (a -x)(c-x> 

37.^' = ^. 38. 2x + l = 3. 
b d X 

33 15x^810^ 40. x2 + ? = 50. 

2 3x 7 

- - 2 X 1050 yio« + a^,^ + a^ 5 

41. — = 4<c. 1 = -. 

3 7x 6 + xa + x2 

43 ^ + ^^ :^ 2x + l 44 x3- 10x2 + 1 ^ ^ _ g 
■x + 3 x+6* 'x2-6x + 9 

.^ ax + b mx — n ac ^^ 1^ 4 

45. = 46. = 0. 

6x + anx — m xx — 2x-3 

47. -— ^— = -. 48. — - — + = X - 1. 



49. 



te2 — mx + n n 9 2 x — 3 

x-2 _ 3 (8 - x) gQ (a-x)3 + (x-6)8 _ a^-b^ 

}x+ 14~ 28-x ' * " 



51.'^— + ^-^ = 12. 52. 

2x- 3 X- I 

53 5 + X ' 8 - 3 X _ 2 X -^ 

' 3 — X X x-2 

55. (^'=8(^-15. 56.5^ 

\x -b) \x-bj 9 

57. a2 - x2 = (a - X) (6 + c - x). 



(a- 


-X)- 


-(X- 


-6) 


a 


+ &■ 


16- 


■ X 


2(x 


-11) 


x-4 


4 




X • 


-6 




12 


2x- 
X — 


T* 


3x- 

X - 


4-1 
-3 


5x 

X 


-14 
-4 


5x- 


^V 


3x- 


-1_ 


?4 


■X — 



58. i= + ^ = ^. 

59 2_^±^ + J?- = ^Ili + ^Z^. 
18 x + 4 4 6 

60. {X - a + 6) (X - a + c) = (a - 6)2 - x^. 



QUADKATIC EQUATIONS 75 

91. Solution of quadratic equations by factoring. When the 
left-hand member of an equation can be factored readily, this is 
the most convenient method of solution. It also illustrates very 
clearly the meaning and property of the roots of the equation. 

Example. Solve x^ -}- 2 x — 15 = 0. 

Factor the left-hand member, (x + 5) (x — 3) = 0. 

The object in solving an equation is to find numbers that substituted for 
the variable satisfy the equation. But since zero multiplied by any number 
is zero (p. 3), any value of x which causes one factor of an expression to 
vanish makes the whole expression vanish. If in this case x = 3, our equa- 
tion in factored form becomes 

(3 + 5) (3 - 3) = (3 + 5) • = 
and is satisfied. If we let x = - 5, the other factor becomes zero, and the 
equation reduces to the identity 

(5_5)(6-3) = 0(5-3) = 0. 

Thus the numbers 3 and — 6 are solutions of the equation. 

92. Solution of an equation by factoring. We have imme- 
diately the 

EuLE. Transpose all the terms to the left-hand member of the 
equation. 

Factor that member into linear factors. 

The values of the variable that make the factors vanish are 
roots of the equation. 

EXERCISES 

Solve and check the following : 

1. 6x2 + x = 15. 

3. 6x2 -X- 6 = 0. 

5. 13x2-38x = 3. 

7. x2-40x + 111 = 0. 

9. x2-18x- 208 = 0. 
11. x2-3ax-4a2 = 0. 
13. (x2 _ 1) + (X - 1)2 = 0. 

15. (2x - l)(x -I- 2) -f (X - 1) (X - 2).= - 4. 

16. (7x - l)(x -f 3) - (4x - 3)(x - 1) =24. 



2. 


6x2-f-7x = 3. 


4. 


5x2-17x+6 = 0. 


6. 


2x2 -6x- 26 = 0. 


8. 


13x2 -40x + 3 = 0. 


10. 


3x2 -26x + 36 = 0. 


12. 


(X - a)2 - (x - 6)2 = 0. 


14. 


(3x_6)2_(9x-M)2 = 0. 



76 QUADRATICS AND BEYOND 

17. (3x + 1) (X + 1) - (4a; + 3) (x - 1) = - 2. 

18. (X - a)(4ax - 6) + (x - 6)(4ax - 6) = 0. 

19. 3x-4Vx^^ = 2(x + 2). 



Solution : Transpose, x — 4 = 4 Vx — 7. 

Square, x2 - 8 x + 16 = 16 x - 112. 

Transpose, x2 - 24 x -f 128 = 0. 

Factor, (x - 16) (x - 8) = 0. 

The roots are x = 16. x = 8. 



Check : 3-16-4 Vl6 - 7 - 2 (16 + 2) = 48 - 12 - 32 - 4 = 0. 
3-8-4 VS~^ - 2(8 + 2) = 24 - 4 - 20 = 0. 

In the following examples, as always, the quadratic equation should be solved 
by factoring when possible. Recourse to the longer but sure method of completing 
the square is always available. 

When an equation is cleared of fractions or squared in the process of bringing 
it into quadratic form (1), §88, extraneous solutions may be introduced. The 
results should be verified in every case and extraneous solutions rejected. 

20. a + Va2 - x^ = x. 21. VxT~5 = x - 1. 



22. 2x-^V2x-l = x + 2. 23. 1- 6x + V5(x + 4) = 0. 

24. Vll-x + \/x-2 = 3. 25. V2X + 1-2 V2x + 3 = 1. 



Hint. Square twice. 



26. Va(x-6) + V6(x-a) = x. 



27. VxTS + V2X-3 = 6. 28. 2 V3 + X - 4 VS^^ = V60. 



29. Vl + ox - Vl - ax = X. 30. V5x- 1 - V8-2x = Vx^. 



31. Vx + 7-V5x-2 = 3. 32. x - 10 = |(x - 1)-V2x-1. 



33. Va2"^^ + V&2 + X = a + &. 34. V4 - x + V6 - x = V9 - 2; 



X 

X \b 



35. (a2-62)(x2+l) = 2(a2 + 62)x. 36. x + 2 a V2(a2 + 62) - x = 3a2 + 62. 

37. xV^:r2 + 2V^T2zzV^H^. 38. -J^-^ + \f-^ = 2- 

gg \^a-\-x-\-Va-x _a .^ Va + Vx _ 2 Vx (x+a)2 
Va + X - Va - X X Va-Vx y/a + Vx a{x-a) 

Hint. Rationalize. 

Va — X 4- Vx — 6 ja — x ^^ ^ , y/b-Va_ 1 Va—Vb 



41. V'J^ + v^ ^ /«E^. 42.^+; 
Va - X - Vx-6 \ X - 6 



V6 Vx Va 



QUADRATIC EQUATIONS 77 

^2 2a-(l + a2)a; ^ 26 + (l + 62)x 

4 



44. 2v^+4-3V2x-3 



Vx + 4 

45. (a - a;)2 + (& _ x)2 = f (a - cc) (6 - x). 

46. a&x2 - (a + 6) (a6 + 1) x + (a6 + 1)2 = 0. 



47. V2x-2+V3x + 7 = V2x+ll + V3x-8. 

48. V{a + x) (X + 6) + V(a - x) (x - b) = 2 Vox. 



49. 2 72 a + 6 + 2 X - Vl0a + 6-6x = Vl0a + 96-6x. 

93. Quadratic form. Any equation is in quadratic form if it 

may be written as a trinomial consisting of a constant term and 
two terms involving the variable (or an expression which may be 
considered as the variable), the exponent in one term being twice 
that in the other. By the constant term is meant the term not 
containing the variable. 

Thus X -S V ^ + 13 = 0, a ;-^ + x"^ - 3 = , a2x-2» _ (a + 6) a;-« + 62 ^ 0, 
x2 — 2x — S— Va;2 _ 2 x — 3 + 17 =^ are all in quadratic form. In the last the 
whole expression x2 _ 2 x — 3 is taken as the variable. 

It is usually convenient to replace by a single letter the lower 
power of the variable or expression with respect to which the 
equation is in quadratic form. 

EXERCISES 

Solve and check the following : 

1. X - 8 v^ + 15 = 0. (1) 

Solution : Let y/x = y. 

Then x = y2. 

Substituting, (1) becomes y2 _ 3^ _j. 15 = 0. 
Factor, (y - 6) (y - 3) = 0. 

The roots are y = 5, y = S. 

Thus Vx = 6, Vx = 3, 

or X = 25, X = 9. 

Check: 25 -8-5 + 15 = 0; 9-8-3 + 15 = 0. 



78 QUADRATICS AND BEYOND 



2. x-i - 


5x-§ + 4 = 


= 0. 




Solution : 


Let 




X-3 = y. 


Then 






X-t = y1. 


Substituting, (1) becomes 


2/2_ 5y + 4 = 0. 


Factor, 






(y - 4) (y - 1) = 0. 


The roots are 




2/ = 4, y = \. 


Since 






x-l-^- ' . 
x3 ^' 


we have ' 






VX2 ^2 


Thus 


4' 


X2 = 


1 1 

= 64' "^-=^8' 


and 


1' 


X2 = 


:1; x=±Vl = ±l. 



(1) 



3. 2Vx2-2x-3 + x2 - 2x = 
Solution : Add — 3 to both members and rearrange terms, 

x2 _ 2x - 3 + 2 Vx2 - 2x - 3 - 3 = 0. (1) 

Let Vx2 - 2 X - 3 = y. 

Then x2-2x -3 = 2/2. 

Substituting, (1) becomes ^2 ^ 2 y — 3 = 0. 

. Factor, (y + 3) (y - 1) = 0. 

The roots are y = —3, y = 1. 

Hence y2 _ ^2 _ 2 x - 3 = 9, 

or x2_2x + l = 13. 

Extract the square root, x — 1 = ± "s/l3. 

The roots are x = 1 ± VlS. 

Also x2 - 2 X - 3 = 1, 

or x2-2x + l = 5. 

Extract the square root, x — 1 = ± V5. 

The roots are x = 1 ± VB. 

4. x8 - 1 = 0. 5. x8 - 8 = 0. 

6. x8-l = 0. 7. x5 + 8xi = 9x. 

8. X - 6 x-i = 1. Hint. Divide by Vx. This factor cor- 

responds to the root z = 0. 

9- 4x3 + 5x^-1 = 0. 10. 10x4-21 = x2. 



QUADRATIC EQUATIONS 79 

11. 2x^-3x^-\-x = 0. 12. 3v^ + 6-l^=4. 

13. ax^p + bxp + c = 0. 14. 3 ic* - 7 x2 - 6 = 0. 

15. 4x6 - 14x3 + 6 = 0. 16. x* - 13x2 + 36 = 0. 

17. x~ 12Vx + ll = 0. 18. x3 + 4x^= 16ixVx2. 

19. ■\/x3-2Vx + x = 0. 20. 8X-6 + 999x-3= 125. 

21. 7^x6 + 5xv^ = 66. 22. 2(V5 - 3)=* - 3 = v^. 

23. (x2 - 10) (x2 - 3) = 78. 24. (x2 _ 1)2 + (a;2 + i) = 2. 

25. ( v^ - 1)^ + v^ = ^z. 26. x^ + x^ = (28 + 2-8)x V^xl 

27. (Vx- 3)(v'x - 4) = 12. 28. (X + a)^ + (x + 6)* = (a - b)K 

29. (2x2-3)2-(x2 + 4)2 = 7. 30. 2x2 + 3 Vx2 - x + 1 = 2x + 3. 

31. (2x + 3)' + (2x + 3)-^:^0. 32. 8(8x - 5)8 + 5(5 - 8x)« = 85. 

33. x4-4(a+6)x2+16(a-6)2 = 0. 34. 4x2 + 12xVTTi = 27(1 + x). 

35. (2x2-3x+l)2=22x2-33x + l. 36. (x2_ 5x+7)2 -(x-2)(x-3) = l. 




40. ^^ = x-8. 



37. x2 + 5 = 8x + 2 Vx2 - 8x + 40. 38. 2 V(x - 4)2 + 4(x - 4)"^ = 9. 

x-4 
2 + x 
42. a = x2 + 6 + -. 

44.1+^- , ^' , =3. 

2 2(l + vT+^)' 

.- 7 , 41 Vx 97 . 5 ._ 2x (x-l)2 . 

45. X* H = h x5. 46. h ^^ = 4. 

X ^^ (x-l)2 2x 

47. (X - 1)2 + 4 (X - 1) - 5 = 0. Hint. I.ety = -^^: then- = ^^~^^'' ' 

^ ' ^ ' " (X — 1)2 y 2x 

48. (x2 + 2)^+ ^ =4x2 + 8. 
Vx2 + 2 

94. Problems solvable by quadratic equations. The principle 
of translating the problem into algebraic symbols, explained in 
§ 55, should be observed here. The result should be verified in 
every case. It may happen that the problem implies restrictions 
that are not expressed in the equation to which the problem 
leads. In this case some of the solutions of the equation may 
not be consistent with the problem ; for instance, when the 



80 QUADKATICS AND BEYOND 

variable stands for a nuinber of men fractional solutions should 
be rejected. If only such results are obtained, the problem is 
seK-contradictory. Often negative solutions should be rejected, 
as when the result indicates a negative number of digits in a 
number. Imaginary or complex (p. 72) results in general mean 
that the conditions of the problem cannot be realized. 

PROBLEMS 

1. The product of ^ and ^ of a certain number is 500. "What is the 
number ? 

2. There are two numbers one of which is less than 100 by as much as 
the other exceeds it. Their product is 9831. What are the numbers ? 

3. The sum of the square roots of two numbers is VtI. One of the 
numbers is less than 37 by as much as the other exceeds it. What are the 
numbers ? 

4. A man sells oranges for -^^ as many cents apiece as he has oranges. 
He sells out for |3. How many had he ? 

5. If the perimeter of a square is 100 feet, how long is its diagonal ? 

6. A man sells goods and makes as much per cent as ^ the number of 
dollars in the buying price. He made $246. What was the buying price ? 

7. Two bodies A and B move on the sides of a right angle. A is now 
123 feet from the vertex and is moving away from it at the rate of 239 feet 
per second. B is 239 feet from the vertex and moves toward it at a rate of 
123 feet per second. At what time (past or future) are they 850 feet apart ? 

8. What is the number twice whose square exceeds itself by 190 ? 

9. What numbers have a sum equal to 63 and a product equal to 612 ? 

10. The sum of the squares of two numbers whose difference is 12 is 
found to be 1130. What are the numbers ? 

11. By what number must one increase each factor of 24 • 20 so that the 
product shall be 540 greater ? 

12. What numbers have a quotient 4 and a product 900 ? 

13. Two numbers are in the ratio of 4 : 5. Increase each by 15 and the 
difference of their squares is 999. What are the numbers ? 

14. If 4^ is divided by a certain number, the same result is obtained as 
if the number had been subtracted from 4^. What is the number? 

15. Separate 900 into two parts such that the sum of their reciprocal 
values is the reciprocal of 221. 



QUADRATIC EQUATIONS 81 

16. The denominator of a fraction is greater by 4 than the numerator. 
Decrease the numerator by 3 and increase the denominator by the same, 
and the resulting fraction is half as great as the original one. What is the 
original fraction ? 

17. The numerator and denominator of a fraction are together equal 
to 100. Increase the numerator by 18 and decrease the denominator by 16, 
and the fraction is doubled. What is the fraction ? 

18. A number consists of two digits whose sum is 10. Reverse the order 
of the digits and multiply the resulting number by the origingll one, and 
the result is 2944. What is the number? 

19. The sum of two numbers is 200. The square root of one increased by 
the other is 44. What are the numbers ? 

20. The difference of two numbers is 10, and the difference of their cubes 
is 20630. What are the numbers ? 

21. Around a rectangular flower bed which is 3 yards by 4 yards there 
extends a border of turf which is everywhere of equal breadth and whose 
area is ten times the area of the bed. How wide is it ? 

22. Two bicyclists travel toward each other, starting at the same time 
from places 51 miles apart. One goes at the rate of 9 miles an hour. The 
number of miles per hour gone by the other is greater by 5 than the number 
of hours before they meet. How far does each travel before they meet ? 

23. A printed page has 15 lines more than the average number of letters 
per line. If the number of lines is increased by 15, the number of letters per 
line must be decreased by 10 in order that the amount of matter on the two 
pages may be the same. How many letters are there on the page ? 

24. A merchant buys goods for a certain sum. The cost of handling them 
was 5% of their cost price. He sells for §504, gaining as much per cent as 
T^Q the cost price was in dollars. What was the cost price ? 

25. A man had $8000 at interest. He increased his capital by $100 at the 
end of each year, apart from his interest. At the beginning of the third year 
he had $8982. 80. What per cent interest did his money draw ? 

26. Two men A and B can dig a trench in 20 days. It would take A 9 
days longer to dig it alone than it would B. How long would it take B 
alone ? 

27. A cistern is emptied by two pipes in 6 hours. How long would it 
take each pipe to do the work if the first can do it in 5 hours less time than 
the second ? 

28. A party procures lunch at a restaurant for $16. If there had been 
5 less in the party, each member would have paid 15 cents more without 
affecting the amount of the entire bill. How many were in the party ? 



82 QUADRATICS AND BEYOND 

29. A party pays $12 for accommodations. Had there been 4 more in the 
party, and if each person had paid 25 cents less, the bill would have been 
$16. How many were in the party ? 

30. A grocer sells his stock of butter for $16. If he had had 6 pounds less 
in stock, he would have been obliged to charge 10 cents more a pound to 
realize the same amount. How many pounds had he in stock ? 

31. A man buys lemons for $2. If he had received for that money 50 
more lemons, they would have cost him 2 cents less apiece. What was the 
price of each lemon ? 

32. It took a number of men as many days to dig a ditch as there were 
men. If there had been 6 more men, the work would have been done in 8 
days. How many men were there ? 

95. Theorems regarding quadratic equations. In this and the 
following sections we prove several theorems concerning quad- 
ratic equations. Similar theorems are later proved in general for 
equations of higher degree. 

Theorem. If a is a root of the equation 

ac(^-\-hx-\-c = 0, (1) 

then X — a is a factor of its left-hand member , and conversely. 

The fact that a is a root of the equation is equivalent to the 
assertion of the truth of the identity 

aa^ -\- ha ■\- G =i Oj 

by definition of the root of an equation (§ 53). 
Divide ax^ 4- Jx -f c by ic — a as follows : 

X — a\ ax^ -\- bx -\- c \ax + (^ 4- ao^) 



(b + aa) X -{- c 

(b -f- aa) X — aa^ — ba 

aa^ -{- ba -\- c 

Since the remainder vanishes by hypothesis, ax^ + bx -{- c is 
exactly divisible hy x — a. 

Conversely, we have already seen (p. 3) that ii x — a is a, 
factor of an equation, a is a root, since replacing x hy a would 
make that factor vanish. 



QUADRATIC EQUATIONS 83 

EXERCISES 

Form equations of which the following are roots. 

1. 2, 6. 

Solution : Since 2 and 6 are roots, (x — 2) and (x — 6) are factors, and the 
equation having these as factors is 

(X - 2) (X - 6) = x2 - 8 X + 12 = 0. 

2. 1, - 1. 3.-3,-4. 
4. V2, - V2. 5. - 2, - 6. 

6. - 4 V2, V32. 7. V27, - 3 VS. 

8. 2 + V3, 2 - V3. 9. a + V6, a - V6. 

96. Theorem. A quadratic equation has only two roots. 

Given the equation ax^ + ?)X + c = with the roots a and /3, to prove that 
the equation has no other root, as 7, distinct from a and /3. 

Since a and j8 are roots of the equation, x — a and x — /3 are factors. 
Thus our equation may be written in the form 

a(x-a:)(x-/3) = 0. (1) 

If now 7 is a root, it must satisfy (1), i.e. 

a(7-a)(7-/3) = 0. 

But in order that any product of numerical factors should vanish one of 
the factors must vanish. Tlius either a = 0, or 7 — or = 0, or 7-/3 = 0. 
But, by hypothesis, 7 5^ <x and 7 ^ /S, so the last two factors cannot vanish. 
Thus a = 0. This would, however, reduce our equation to a linear equa- 
tion, which is contrary to our hypothesis that the equation is quadratic. 
Thus the assumption that we have a third distinct root leads to a contradiction. 

CoKOLLARv I. If a quadratic equation is satisfied by more than two distinct 
values of the variable, then each of the coefficients vanishes identically. 

The above proof shows that the coefficient of x^ must vanish. In the same 
way it can be shown that b = c = 0. 

Corollary II. If two quadratic expressions have the same value for more 
than two values of the variable, then their coefficients are identical. 

Let ax2 + 6x + c = a'x^ + b'x + c' 

for more than two values of x. Transpose, and we obtain 
(a - a')x^ + (6 - b')x + c - c' = 0. 

We have then a quadratic equation satisfied by more than two values 
of X. Thus by Corollary I each of its coefficients must vanish. Thus a' = a, 
6' = 6, c' = c. 



84 QUADRATICS AND BEYOND 

This theorem taken with § 95 is equivalent to the statement that a quad- 
ratic equation can be factored in one and only one way. Thus if 

ax"^ + 6x 4- c = a (x — or) (x — /3), 
we cannot find other numbers 7 and 5 distinct from a and /3 such that 

ax2 + 6x + c = a (x — 7) (x — 5), 
for then the equation would have roots distinct from a and /3. 

97. Theorem. If the equation 

x2 + 6x + c = 0, (1) 

where b and c are integers, has rational roots, those roots must be integers. 

P 
For suppose - to be a rational fraction reduced to its lowest terms and 

a root of (1). 



Then 


^, + ^-^ + = 0. 
q2 q 




or 


p2 + ipq + c^2 = 0, 




which gives 


p'^ = - q{bp-{- cq). 




Thus some 


factor of q must be contained in p (§ 69), which contradicts 


the hypothesis 


that the fraction - is already reduced to its lowest terms. 


98. Nature of the roots of a quadratic equation. 


The equation 




ax^ -^ bx -\- c = 


(1) 



has as roots (§ 89) x^ = "^ ^ ^^, (2) 



_ ^ -l_ V^'^ - 


- 4c ac 


2a 




-b-^b^- 


- 4: ao 



2a ■ (^) 

These expressions afford an immediate arithmetic means of 
determining the nature of the roots of the given equation when 
a J b, and c have, numerical values and a =^ 0. In fact an inspec- 
tion of the value of 6^ — 4 ac is sufficient to determine the nature 
of the roots of (1). 'mM 

I. When h^ — Jf.ac is negative, the roots are iikhg%naty (§89). 
II. When y^ — Ji^ac^Oy the roots are real and equal. In this 

b 
case x^=x^=-—- 

III. When y^ — Ji^ac is positive, the roots are real and distinct. 

IV. When ¥ — Jiac is positive and a perfect square, the roots 
are real, distinct, and rational. 



QUADRATIC EQUATIONS 85 

In case IV, if J^ — 4 ac = A, 

_ — h + Va _ — ^, - Va 

^' ~ 2 a ' ^"^ " 2 a ' 

The converses of these four cases are also true. For instance, 
if the roots of (1) are imaginary, from (2) and (3) it is clear 
that P — 4: ac must be negative. 

The expression A = Z>^ — 4 ac is called the discriminant of the 
equation ax^ -\- bx -^ c = 0. 

EXERCISES 

1. Determine the nature of the roots of the following equations without 
solving. 

(a) 3x2 _4x-l =0. 

Solution : A = (- 4)2 - 4 . 3 . (- 1) = 16 + 12 = 28 and is then positive. 
Thus by III the roots are real and distinct. 

(b) 3x2 - 7x + 6 = 0. (c) 6x2 - X - 1 = 0. 
(d) 3x2 + 4x + l = 0. (e) x2-4x + l = 0. 
(f) 2x2 -6x-9=:0. (g) 2x2-4x-2 = 0. 
(h) 4x2 + 12x + 9 = 0. (i) 2x2 + 6x - 4 = 0. 

(j) 4x2- 28x + 49 = 0. (k) 4x2 + 12x + 5 = 0. 

2. Determine real values of k so that the roots of the following equations 
may be equal. Check the result. 

(a) (2 + A;)x2 + 2A:x + l = 0. 

Solution : Here 2 -\- k = a, 2fc = 6, l = c. 

Thus A = 62-4ac = 4fc2_4.(fc + 2).l 

= 4 ^2 _ 4 A: _ 8. 
Since the roots of an equation are equal when and only when its dis- 
criminant equals zero (§ 98, II), the required values of k make A = and 

are the roots of 

4^2 _ 4^ _ 8 = 0, 

or fc2 - A: - 2 = 0. 

Solve by factoring, 

k^-k-2=:{k-2){k + l) = 0. 
Thus the values of k are /c = 2, k = — 1. 

Check : Substituting in the original equation for /c = 2, we get 
4x2 + 4x + l = (2x + l)2, 
and for A: = — 1 we get x2 — 2 x + 1 = (x — 1)2. 



86 QUADRATICS AND BEYOND 

(b) x2 + fcx + 16 = 0. (c) a;2 + 2a; + A;2 = 0. 

(d) x2-2A:x + l = 0. (e) 3A;x2-4x-2 = 0. 

(f) A;x2-3x + 4 = 0. (g) x^ -\- 4kx -\- k^ + 1 = 0. 

(h) A:2x2 4- 3x - 2 = 0. (i) (A:2 + 3) x2 + fee - 4 = 0. 

(j) 3/cx2 + A-x - 1 = 0. (k) x2 + (3A; + l)x + 1 = 0. 

(1) x2 + 3x + A; - 1 = 0. (m) x2 + 9A;x + 6fc + ^ = 0. 

(n) 4 A;2x2 + 4 A-x - 125 = 0. (o) 2 x2 - 4x - 2 A; + 3 == 0. 

(p) (A: + l)x2 + A:x + A; + 2 = 0. (q) A;x2 + (4 A; + l)x + 4 A: - 3 = 0. 

(r) 2(A:4- l)x2 + 3A:x + A;-1 = 0. (s) (A: - l)x2 + 5A:x + 6A; + 4 = 0. 

(t) (2A; + 3)x2-7A;x+^^^ = 0. (u) (A;-l)x2+(2A; +l)x+ A: + 3 = 0. 



CHAPTEE IX 
GRAPHICAL REPRESENTATION 

99. Representation of points on a line. Let us select on the 
indefinite straight line AB sl certain point as a point of refer- 
ence. Let us also select a certain line, the length of which for 
the purpose in hand shall represent unity. Let us further agree 
that positive numbers shall be represented on ^J5 by points to 
the right of 0, whose distances from are measured by the given 

-3 -2 -1 +1 +2+3 1 



numbers, and negative numbers similarly by points to the left. 
Then there are certainly on AB points which represent such num- 
bers as 2, — 3, ^, — 1^, or, in fact, any rational numbers. Since 
we can divide a line into any desired number of equal parts, we 
are able to find by geometrical construction the point correspond- 
ing to any rational number. Furthermore, by the principle that 
the square of the hypotenuse of a right triangle equals the sum 
of the squares of the other two sides, we can find the point 
corresponding to any irrational number expressed by square-root 
signs over rational numbers. More complicated irrational num- 
bers cannot, however, in general be constructed by means of ruler 
and compasses, but we assutne that to every real number there 
corresponds a point on the line, and conversely , we assert that to 
every point on the line corresponds a real number. This assump- 
tion of a one-to-one correspondence between points and real num- 
bers is the basis of the graphical representation of algebraic 
equations. 

This amounts to nothing more than the assertion that every real number, 
rational or irrational, as, for instance, — 6, 2 + V3, V3, tt, represents a certain 
distance from on AB, and conversely, that whatever point on the line we may 
select, the distance from to that point may he expressed by a real number. . 

87 



88 



QUADRATICS AND BEYOND 



'<Y 



X axis 



T^ 



100. Cartesian coordinates. We have seen, that when the 
single letter x takes on real values all these values may be repre- 
sented by points on a straight line. 
When, however, we have two variables, 
as X and y, which we wish to represent 
simultaneously, we make use of the 
plane. Just as we determined arbitra- 
rily, on the line along which the single 
variable was represented, an arbitrary 
point for the point of reference and an 
arbitrary length for the unit distance, 
so now we select an indefinite line along which x shall be repre- 
sented, and another perpendicular to it along which 7/ shall be 
represented. The former we call the X axis ; the latter the Y axis. 
The intersection of these axes we take as the point of reference 
for each. This point is called the origin. 

We select a unit of distance for x and a unit of distance for 1/ 
-v^hich may or may not be the same, according to the problem 
under discussion. As before, we represent positive numbers on 
the X axis to the right, and negative numbers to the left. Positive 
values of y are represented above the X axis, and negative values 
below it. The arrowhead on the 
axes indicates the positive direc- 
tion. Any pair of values of x 
and y, written (x, ?/), may now 
be represented by a point on 
the plane which is x units from 
the Y axis and y units from the 
X axis. Thus if x = 0, y = 0, 
written (0, 0), the point repre- 
sented is the origin. The point 
(3, 0), i.e. a; = 3, ?/ = 0, is found 
by going three units of x to the 
right, i.e. in the positive direction of x and no units up. The point 
(4, 3), i.e. a = 4, 2/ = 3, is found by going four units of x to the 
right and three units of y up. The point (— 3, 1) is found by 



3,4) 



YA 



0(0,7) 



U',3. 



3.0) 



{4, -2) 



GRAPHICAL REPRESENTATION 89 

going three units of x to the left and one unit of y up. The point 
(— 3, — 4) is found by going three units of x to the left and four 
units of y down. In fact, if we let both x and y take on every 
possible pair of real values, we have a point of our plane corre- 
sponding to each pair of values of (x, y). Conversely, to every 
point of the plane correspond a pair of values of (x, y). These 
values are called the coordinates of the point. The value of a?, 
i.e. the distance of the point from the Y axis, is called its 
abscissa; the value of y, i.e. the distance of the point from the 
X axis, is called its ordinate. If the point (x, y) is conceived as a 
moving point, and if no restriction is placed . 

on the value of the coordinates so that they 

^ ., T 'PIT ^>^<^ Quadrant ist Quadrant 

take on every possible pair oi real values, (-,+; (-|-,+) 



ith Quadrant 



every point in the plane is reached by the ^ 

moving point (X, y). 3rd Quadrant 

The X and Y axes divide the plane into (-'-) 
four parts called quadrants, which are num- 
bered as in the figure. The proper signs of the coordinates of 
points in each of the quadrants are also indicated. 

EXERCISES 

The following exercises should be carefully worked on plotting paper, 
which can be bought ruled for the purpose. 

1. Plot the points (2, 3), (0, 4), (- 4, 0), (- 9, - 2), (2, - 4). 

2. Plot with the aid of compasses the points (l, V'2), (V3, — V2), 
(2+V3, 2-V3),(-V2, -V2). 

3. Plot the square three of whose vertices are at (— 1, — 1), (— 1,+ 1), 
(+ 1, — 1). What are the coordinates of the fourth vertex? 

4. Plot the triangle whose vertices are (2, 1), (—6, — 2), (—4, 4). 

5. Plot the two equilateral triangles two of whose vertices are (6, 1), 
(—6, 1). Find coordinates of the remaining vertices. 

6. If the values of the coordinates (x, y) of a moving point are restricted 
so that both are positive and not equal to zero, where is the point still free 
to move ? 

7. If the coordinates {x, y) of a moving point are restricted so that con- 
tinually y = 0, where is the point still free to move ? 



90 



QUADRATICS AND BEYOND 



8. What is the abscissa of any point on the T axis ? 

9. The coordinates of a variable point are restricted so that its ordinate 
is always 2. Where may the point move ? 

10. If both ordinate and abscissa of a point vanish, can the point move ? 
Where will it be ? 

11. Plot the quadrilateral whose vertices are (0, 0), (— 6, — 3), (5, — 5), 
(—1, — 8). What kind of a quadrilateral is it? 

12. The coordinates of three vertices of a parallelogram are (— 1, — 1), 
(6, 2), (—1, — 6). Find the coordinates of the fourth vertex. 

13. The coordinates of two adjacent vertices of a square are ( — 1, — 2) and 
(1, — 2). Find the coordinates of the remaining vertices (two solutions). Plot 
the figures. 

14. The coordinates of two adjacent vertices of a rectangle are (— 1, — 2), 

(1, — 2). What restriction is imposed on the coordinates of remaining 
vertices ? 

15. The coordinates of the extremities of the bases of an isosceles triangle 
are (1, 6), (1, — 2). Where may the vertex lie? What restriction is imposed 
on the coordinates (x, y) of the vertex ? 

101. The graph of an equation. The equation a? = 2 ?/ is satis- 
fied by numberless pairs of values (x, ?/); for example, (2, 1), 
(0, 0), (1, ^), (— 2, — 1) all satisfy the equation. There are, how- 
ever, numberless pairs of values which do not satisfy the equation ; 

for example, (1, 2), (2, - 1), (- 1, 1), 
(0, — 1). The pairs of values which 
satisfy the equation may be taken 
as the coordinates of points in a 
plane. The totality of such points 
would thus in a sense represent the 
equation, for it would serve to dis- 
tinguish the points whose coordi- 
nates do satisfy the equation from 
those whose coordinates do not. 
After finding a few pairs of values which satisfy the above equa- 
tion we note that any point whose abscissa is twice the ordinate, 
i.e. for which cc = 2 ?/, is a point whose coordinates satisfy the 
equation. Any such point lies on the straight line through the 
origin and the point (2, 1). We can then say that those points 



I 




GRAPHICAL REPRESENTATION 91 

and only those which are on the straight line represented in the 
figure have coordinates which satisfy the equation. This line is 
the graphical representation or graph of the equation. 

The equation of a line or a curve is satisfied by the coordinates 
of every point on that line or curve. 

Any point whose coordinates satisfy an equation is on the 
graph of the equation. 

102. Restriction to coordinates. Iri § 100 it was seen that a 
moving point whose coordinates were unrestricted took on every 
position in the plane. We now see that when the coordinates of 
a point are restricted so as to satisfy a certain equation (as x = 2 3/), 
the motion of the point is no longer free, but restrained to move 
along a certain path. Thus, for instance, the equation a; = 4 means 
that the path of the moving point is so restricted that its abscissa 
is always 4. Its ordinate is still unrestricted and may have any 
value. This shows that the plot of a? = 4 is a straight line four 
units to the right of the Y axis and parallel to it, for the abscissa 
of every point on that line is 4, and every point whose abscissa is 
4 lies on that line. 

EXERCISES 

Determine on what line the moving point is restricted to move by the 
following equations. Draw the graph. 

1. x = 6. 2. x = 0. 3. ?/ = f. 

4. X = y. h. y = 2. 6. 3 X = y. 



7. 2x = y. 8. y = 0. 9. x = -3. 

10. X = 32/. 11. 3y = - X. 12. X + y = 0. 

13. 6x = ll. 14. 2/=-3. 15. 2x = -3?/. 

16. x = -2y. 17. x = -l. 18. 2x-62/ = 0. 

103. Plotting equations. Plotting an equation consists in find- 
ing the line or curve the coordinates of whose points satisfy the 
equation. Thus the process of § 101 was nothing else than plot- 
ting the equation x = 2y. This may be done in some cases by 
observing what restriction the equation imposes on the coordinates 
of the moving point ; but more often we are obliged to form a 



92 



QUADRATICS AND BEYOND 




table of various solutions of the equation, and to form a curve 
by joining the points corresponding to these solutions. This 
gives us merely an approximate figure of the exact graph which 
becomes more accurate as we find the coordinates of points closer 
to each other on the line or curve. 

EuLE. Wlien y is alone on one side of the equation, set x equal 
to convenient integers and compute the corresponding values of y. 

Arrange the results in tabular form. Take corresponding 
values of x and y as coordinates and plot the various points. 

Join adjacent points, making the entire plot a smooth curve. 

When X is alone on one side of the equation integral values of y may be assumed 
and the corresponding values of x computed. 

Care should be taken to join the points in the proper order so that the resulting 
curve pictures the variation of y when x increases continuously through the values 
assumed for it. By adjacent points we mean points corresponding to adjacent 
values of x. 

Any scale of units along the X and Y axes that is convenient may be adopted. 
The scales should be so chosen that the portion of the curve that shows considerable 
curvature may be displayed in its relation to the axes and the origin. 

When there is any question regarding the position of the curve between two 
integral values of x, an intermediate fractional value of x may be substituted, the 
corresponding value of y found, and thus an additional point obtained to fix the 
position of the curve in the vicinity in question. 



EXERCISES 
Plot: 

1. x2 -4a; + 3 = y- 

Solution : In this equation if we set x = 0, 
1, 2, 3, etc., we get 3, 0, — 1, 0, etc., as cor- 
responding values of y. Thus the points 
(0, 3), (1, 0), (2, - 1), (3, 0), etc., are on 
the curve. These points are joined in order 
by a smooth curve. 



X 


y 


X 


y 





3 


-1 


8 


1 





-2 


15 


2 


-1 




3 







4 


3 




6 


8 




6 


16 







\ 


W^ 














1/ 




1 
























































































































































































i 










\ 










/ 












\ 










/ 












\ 










/ 












' 










/■ 




















i 
















\ 






/ 
















\ 






f 



















\ 


/ 










*x 






— 


— 






— 








- 



GRAPHICAL REPRESENTATION 



93 



2. y = x^-lx + l. 
4. y = x^ - Sx + 2. 
6. y = x^ — 2x + 1. 
8. 2/ = 2x2-6x + 7. 
10. y = 2x^-6x-3. 



3. y = x2 + 1. 

5. y = x^ — 4x. 

7. y = x'^ + 6x + 5, 

9. y = 2x2^-3x + 4. 

11. 2/ = x2-12x + ll. 



104. Plotting equations after solution. When neither x nor y 
is abeacly alone on one side of the equation, the equation should 
be solved for y (or x) and the rule of the previous section applied. 
It should be noted that when a root is extracted two values of 
y may correspond to a single value of x. 



EXERCISES 



Plot: 

1. 2x2+ 3y2 = 9. 

Solution 



32/2 = 9 
2/2 = 3 



2x2, 



3/ = ± V3-fx2. 
Assuming the various integral values for x, we obtain the following 



table and plot: 



X 


y 





±V3 = ±1.7 


1 


±VI = ±i-5 


2 


±Vi=± -67 


+ 3 


imaginary 



X 


y 


-1 
-2 
-3 


±VJ = ± -57 
imaginary 



N: 




In this example, when x is greater than 3 or less than — S, y is imaginary. 
Thus none of the curves is found outside a strip x = ±S. 

To find exactly where the curve crosses the X axis, the equation may be 
solved for x and the value of x corresponding to y = found. Thus 



X = ± VF^I^'- 

If 2/ = 0, X = ± Vl = 2-1. This point is included in the plot 
2. X2/ = 4. 



Solution : 



4 



94 



QUADRATICS AND BEYOND 



Form table for integral values of x. 



i 



X 


y 


-1 


-4 


-2 


-2 


-3 


-f 


-4 


-1 


-6 


-1 


-8 


-\ 


-12 


-i 



Since this table does not give us any 
idea of the curves between -\- 1 and — 1, 
v^e supplement the table by assuming 
fractional values for x. 



3. x^ = y\ 
5. xy = — 1. 

7. x2 + ?/2 = 16. 

9. a;2 + 2/2 = 25. 

11. 2x2/ + 3x = 2. 

13. xy + 2/2 = 10. 

10 — 2/2 



Hint, a; = 



y 



4. X2/ = 16. 

6. X2/ = X + 1. 

8. x2 - 2/2 = 9. 

10. x2 + X = 12 2/. 

12. x2 + 9 2/2 = 36. 

14. X — 2/ + 2 xy = 0. 

2/ 

Hint. x = — -• 

1 + 22/ 



2/ 


X 


6 


-1 


8 


-1 


12 


-i 



y 



-8 
12 



15. 6x2+2x + 32/2 = 0. 16. x2 + 2x + 1 = y2 _ 3^. 

105. Graph of the linear equation. The intimate relation 
between the simplest equations and the simplest curves is given 
in the following theorems. 

Theorem I. The graph of the equation y = ax is the straight 
line through the origin and the point (1, a), where a is any real 
number. 

The proof falls into two parts. 

First. Any point on the line through the origin and the point 
(1, a) has coordinates that satisfy the equation. Let P (Figure 1) 
with coordinates (x', y') be on the line OA. By similar triangles 



V 



or 



ax' 



GRAPHICAL KEPRESENTATION 



95 



Thus the coordinates of any point on the line satisfy the equation. 



C«'y) 





Figure 1 



Figure 2 



Second. Any point whose coordinates satisfy the equation lies 
on the line. 

Let the coordinates (x\ y') of the point P (Figure 2) satisfy the 
equation. Then we have 



or 



y = ax', 

^, = a. 

x' 



Let the ordinate y' cut the line at B. Then by the first part 
of the proof BC = ax', 

EC 



or 



x' 



a. 



Thus 

Hence P lies on the line. 



v' EC 
a=—, = — TJ or ti' = EC. 
x' x' -^ 



Theorem II. The graph of any linear equation in two vari- 
ables is a straight line. 

The general linear equation 

Ax-{-By-{-C = (1) 

may be written in the form 

y = ax-\-b, (2) 

A C 

where a = — — and b= — —f provided ^ ^ (§ 7). It E = 0, 
B E 

the equation Ax -\- C = may be put in the form 

C 



96 



QUADRATICS AND BEYOND 




provided A =^ 0. This is evidently the equation of a straight 
line parallel to the Y axis (§ 102). li B = and A = 0,-we have 
no term left involving the variable. Thus the only case for 

which the theorem demands proof is 
when B ^ 0, and the equation may be 
reduced to form (2). By Theorem I 
we know that the graph oi y = ax is 
a straight line. If, then, we add to 
every ordinate y of the line y = ax the 
constant h, the locus of the extremi- 
ties of the lengthened ordinates will 
lie in a straight line, as one can easily 
prove by Geometry. But any point (cc, y) on the upper line is 
such that its ordinate y is equal to the ordinate of the lower 
line, i.e. ax, and in addition the constant h ; that is, y = ax -\- h. 
Also, since the upper line is the locus of the extremities of the 
lengthened ordinates, every point whose coordinates satisfy the 
equation y = ax + h i^ on this upper line. Thus the equation (1) 
has a straight line as its graph. 

CoEOLLARY. Two lines vjhose equations are in the form 

y = ax-\-h, (3) 

yz=ax-\-V (4) 

are 'parallel. ^ 

For the value of the ordinates of (3) corresponding to a given 
abscissa, say x^^ is obtained from the ordinate of (4) corresponding 
to the same abscissa by adding the constant h — h\ Thus each 
point on (3) is always found h — V units above (below if h — b' 
is negative) a point of (4). Thus the lines are parallel. 

106. Method of plotting a line from its equation. Since the 
equation y = ax-\-h\^ satisfied by the values (0, h), the graph cuts 
the Y axis h units above (below if b is negative) the origin. Since 
it is satisfied by the values (1, a -\-b), the graph passes through 
the point reached by going one unit of x to the right of (0, b) and 
a units up (down if a is negative). These two points determine 
the line. We may then plot a linear equation by the following 



GRAPHICAL REPRESENTATION 



97 



Rule. Reduce the given equation to the form 
y — ax-\-h. 

Plot the point (0, h) as one of the two points that determine 
the line. 

From this point go one unit of x to the right and a units 
of y up {down if a is negative) to find a second point that lies 
on the line. 

Draw the line through these two points. 



EXERCISES 



Plot: 




1. 6x + 2y-5 = 0. 




Write in the form 


y = ax + h, 


and we have 


y=_3x + f. 




a =-3; 




6 = f. 



Plot the point (0, |). 

From this point go one unit of x to the right and three 
units of y down to find the second point, which helps 
determine the line. 






2. 6i 



32/ + ll = 0. 

11 









Yk 
















J 


rV 










/ 












/ 


io. 


n 










/ 








/ 














/ 












/ 


f 









i 




/ 










' 



3. x-y = 0. 

5. X + y = 4. 

7. 2 X - y = 4. 

9. Sx-y = 0. 

11. x-8y = 16. 

13. x = 8(2-2/). 

15. X - y - 1 = 0. 

17. x + 2/ + l = 0. 

19. 12x-3y = l. 



4. x — y — 5 = 0. 

6. 2x = 6(l-y). 

8. 12x + 10y = 5. 
10. 16x-10y = 4. 
12. 2x + y + 3 = 0. 
14. 2x-6y-l = 0. 
16. 2x-2y-6 = 0. 
18. 3x-6y-4 = 0. 
20. 7x-8y-9 = 0. 



107. Solution of linear equations, and the intersection of their 
graphs. The process of solving a pair of independent linear 
equations consists in finding a pair of numbers (x, y) which 
satisfy them both. Though each equation alone is satisfied by- 
countless pairs of values (x, y), we have seen that there is only one 
pair that satisfy both equations. Since a pair of values which 



98 



QUADRATICS AND BF.YOND 



satisfy an equation are the coordinates of a point on its graph, it 
appears that the pair of values that satisfy simultaneously two 
equations are the coordinates of the point common to the graphs 
of the two equations, that is, the coordinates of the points of 
intersection of the two lines. 



EXERCISES 

Find the solutions of the following equations algebraically. Verify the 
results by plotting and noting the coor- 
dinates of the point of intersection. 



1 3^- 

^' 3x- 


-iy 


+ 16: 


= 0, 


-y - 


-7 = 




Solution 








3x- 


'4.y 


+ 16 = 





3x- 


y 


- 7 = 







Sy 


-23 = 


. 






y = 


¥• 


Substituting 


in (2), 








3x = 


7 + ¥ 






X = 





(1) 

(2) 



¥• 



To plot (1) and (2) we get the equa- 
tions in the form y = ax + h and apply 
the rule. Thus 

2/ = fx + 4. 

2/ = 3x-7. 





S'^^'iM 




/\l 




^ \ 




i^'f)'! 


y 


"{0.4)' i 


^^ 


/~ 


z 


/ 


y 


- i I 


"7^ 


t X 








. 7 




r 




K'.-i) 




'-t 






. 


JCo.r) 



2 2x + 3y = 6, 
7x + y = 2. 



5. 



3x-2y = l, 
3x + 2?/ = 5. 



g 2x + y = 3, 
• 8x-72/ = l. 

11 x + y = -i, 
' 4x-3y = 5. 



14. 



17. 



X + y = 5, 
4x-2y = 28. 

x-y = -4, 
2x + 6y = 16. 



6. 

9. 
12. 
15. 
18. 



X + y = 5, 


4 


x-3y = -7, 


3x + y = l. 




4x + y = ll. 


3x-7y = 9, 


7. 


X + 2/ = - 7, 


X + 2 y = 3. 


2x-3y = 6. 


2x-5y = 0, 
x-y = S. 


10. 


X + 2 y = - 10, 
2x-y = 0. 


x-y = 1, 


13. 


x-y = l, 


2x-8y = 3. 


2x-42/ = -16. 


X - 2/ = 2, 


16. 


6x-5y = 5, 


4x- 52/ = 9. 


2 X + 3 y = — 20. 


3x + 2|/ = 9, 
8x-2/ = 2. 


19. 


2x + 6y = -20, • 
3x + y = 2. 



GRAPHICAL REPRESENTATION 99 

108. Graphs of dependent equations. We have defined (§ 57) dependent 
equations as those that are reducible to the same form on multiplying or 
dividing by a constant. Thus two dependent equations are reducible to the 
same equation of the form y = ax + b. Hence dependent equations have as 
their graphs the same straight line. We see now the geometrical meaning of 
the statement that dependent equations have countless common solutions. 
Since their graphs have not one but countless points in common, being the 
same line, it is clear that the coordinates of these countless points will 
satisfy both equations. 

109. Incompatible equations. By our definition (§ 60) incompatible equa- 
tions have no common solution. Since every pair of distinct lines have a 
common point unless they are parallel, we can foresee the 

Theorem. Incompatible equations have parallel lines as graphs. 

Let the equations 

ax -{- by -{- c = 0, (1) 

a'x + b'y + c' = (2) 

be incompatible. This is true (§ 60) when and only when 

ab' - a'b = 0. (3) 



(4) 
(5) 

This may be done if neither 6 nor ¥ equals zero. If both b and b' vanish, 
the lines (1) and (2) are both parallel to the Y axis and hence to each other, 
which was to be proved. But if only one of them vanishes, say & = 0, then 
by (3) a = (§ 5), in which case (1) does not include either variable. Thus 
we may assume that neither b nor 6' vanishes and that (4) and (5) may be 
obtained from (1) and (2). 

By (3) a _a/ 

b~V 

Our equations (4) and (5) become 

a c 

a c' 
y = --x_-, 

which represent parallel lines, by the Corollary, p. 96. 

This theorem completes the discussion of the graphical representation of 
the possible classes (§ 61) of pairs of linear equations. 



Let us then assume (3). 




Write (1) and (2) in the form 




y = 


a c 

—b'-b 


y = 





100 QUADRATICS AND BEYOND 



EXERCISES 



Plot and solve : 




1. 


8x + 2y = 3, 


4x + y = 8. 




3. 


10x-5y = 
2 X - y = 3. 


15, 


5. 


x-1y = l, 




4x-2Sy = 


56. 


7. 


x-Sy = 2, 




6x-lSy = 


36. 



2x + 6y = 1, 
x-^Sy = 1. 

2x-Sy = 6, 
8x- 12y = 24. 

12x-6y=18, 
2x-y=l. 

2x-S + y = 0, 
4x -7 -\-2y =0. 



110. Graph of the quadratic equation. Let 

y = ax^ -{- bx -\- c, (1) 

where as usual a, b, and c represent integers and a is positive. 

If we let X take on various values, y will have corresponding 

values and we may plot the equation as in § 103. A root of the 

quadratic equation 

ax^-}-bx + c = (2) 

is a number which substituted for x satisfies the equation, that 
is, gives the value y = in (1). Thus the points on the graph 
of (1) which represent the roots of the equation (2) are the 
points for which y = 0, that is, where the curve crosses the 
X axis. The numerical value of the roots is the measure of 
the distance along the X axis from the origin to the points where 
the curve cuts the axis. Since this distance is always a real 
distance, only real roots are represented in this manner. 

Theorem. If the graph of (1) has no point in common with 
the X axiSf the equation (2) has imaginary roots, and conversely. 

Every equation of form (2) has two roots either real or imagi- 
nary (§ 89). If the graph of (1) has no point in common with 
the X axis, there is no real value of x for which y = 0, i.e. no 
real root of (2). The roots must then be imaginary. 

Conversely, if (2) has only imaginary roots, there is no real 
value of X which satisfies it, i.e. which makes y = in (1). 
Thus the curve has no point in common with the X axis. 



GRAPHICAL REPRESENTATION 1,01 

This suggests the following universal 

Principle. Non-intersection of graphs corresponds to imagi- 
nary or infinite-valued solutions of equations. 

111. Form of the graph of a quadratic equation. Consider the 

equation ^ o rr ^ 

^ y ^2x^+1 x-\-2. ^ (1) 

By substituting for x a very large positive or negative number, 
say X =± 100, y is large positively. Thus for values of x far to 
the right or left the curve lies far above the X axis. If we 
assign to t/ a certain value, say 3/ = 2, we can find the correspond- 
ing values of x by solving a quadratic equation. Thus in (1) let 

or . 2x^ + 1 x = 0. 

The roots are x^=— 3^, x^ = 0. 

Hence the points (— 3^, 2) and (0, 2) are on the curve (§ 101). 
That is, if we go up two units on the Y axis, the curve is to be 
found three and one half units to the left and also again on the 
Y axis. If in (1) we let y = — 4, the corresponding values of x 
are very nearly equal to each other (— 1^ and — 2), which means 
that the curve meets a line parallel to the X axis and four units 
below it at points very near together. The question now arises, 
Where is the bottom of the loop of the curve ? This lowest point 
of the loop has as its value of y that number to which correspond 
equal values of x. Hence we must determine for what value of 
y the equation (1), that is, the equation 

2x^-{-lx + {2-y)=0, 

has equal roots. Comparing with the equation ax"^ + Jx + c = 



or 



Z - 


= a, 7 = 


--b,2- 


-y=-c. 




Thus the condition h 


2-4ac 


= becomes 






49-4 


•2(2- 


-2/)=0, 




y = 


49- 
8 


16_ 


33 

" 8 


4i. 



102 



QUADRA'EICS AND BEYOND 



I as the corre-' 



Substituting this value of y in (1), we get 
spending value of x. 

This gives a single 
value of y for which 
the values of x are 
equal; hence the graph 
of (1) is a single fes- 
toon as in the figure. 

If we take the gen- 
eral equation 

ax^ -{- bx -{- c = y, 

we find precisely similarly that the bottom of the loop is at a 
point whose ordinate is 

P — Aac A 

y = - 



y 


X 




2 

-4 


-.3+ or -3.2 + 
or - ^ 
- li or - 2 
-If 






4a 



4a 



Thus we see again that if the discriminant is negative the graph 
is entirely above the X axis and both roots are imaginary (§§ 98, 
110), since the ordinate of the lowest point of the loop is positive. 
If the discriminant is positive, the graph cuts the X axis and both 
roots are real. 

The results of this section enable us to determine a value of 
y from the coefficients which determine the lowest point of the 
loop of the curve precisely, and hence to show beyond question 
from the graph whether the equation ^as real or imaginary roots. 



EXERCISES 

Plot the following equations and determine by measurement the roots in 
case they are real. Find in each case the lowest point on the loop. 



1. x2 + X + 1 = y. 

4. x2 + 7 aj + 6 = y. 

7. x2-6x + l = y. 

10. x2 + 2x-l = 2/. 

13. 2x2-x-3 = y. 



3. x2-6x + 10 = ?/. 
6. 3x2-7x-6 = y. 
9. 2x2-9x + 7 = y. 



2. x2 - 4 X + 7 = y. 

5. x2-6x + 9 = y. 

8. x2-6x + 5 = y. 

11. x2 -4x + 4 = y. 

14. 3 x2 + 8 X + 5 = y. 

16. What is the characteristic feature of the plot of an equation whose 

roots are equal ? 



12. 3: 



4x-3 = y. 



15. 4x2+12x+9 = y. 



* 



GRAPHICAL REPRESENTATION 103 

112. The special quadratic adc^ + 6ic = O. When in the quad- 
ratic equation 

ax^ -\-hx^ G = 0, (1) 

c = 0, we can always factor the equation into 

ax^ -\- bx = X (ax -j- b) = 0, 



(—:)=»■ 



or 

Thus the roots are £Ci = 0, X2= 

Conversely, if a; '= is a root, then (§ 95) ic — 0, or x, is a factor 
and the equation can have no constant term. 
This affords the 

Theorem. A quadratic equation has a root eqtcal to zero 
when and only when the constant term vanishes. 

We show in a similar manner that both roots of the equation 
(1) are zero when and only when i = c = 0. 

EXERCISES 

1. Prove the theorem just given by considering the expressions for the 
roots in terms of the coefficients (§ 89). 

2. For what real values of k do the following equations have one root 
equal to zero ? 

(a) x2 + 6x - A: + 1 = 0. (b) 2x2 - 3x + A;2 - 1 = 0. 

(c) x2 + 6a; ^. ^2 ^. 1 ^ 0. (d) 2x2 - 4x + fc2 _ 3^ ^ 0. 

(e) 2x2 + 2fcx-2A;2-4A;-2=0. (f) 6x2 - 4x + 2 A;2 + fc + 7 = 0. 

3. What is the characteristic feature of the plot of an equation which has 
one root equal to zero ? 

4. For what real value of A: will both roots of the following equations vanish ? 

(a) - + 3x - 1 = 0. (b) x2 + (fc2 + 3)x + fc - 3 = 0. 

(c) x2 + (fc2 + l)x + l = 0. (d) x2+(fc-3)x + 2fc2-5fc-3 = 0. 

(e) x2 + (A: + l)x + /c2 -1 = 0. (f) (fc-3)x2 + (fc2_9)x + A;2-4A; + 3=a 



104 QUADRATICS AND BEYOND 

113. The special quadratic ax^ + c = O. This equation may 
be written in the form x^ -\- - = and factored * immediately into 



(^-*-^R)(^-^R)=o> 



which shows that the roots are equal numerically but have oppo« 
site signs. The roots are 



Xi =\/-^' ^2 



xR 



Since in the equation ax^ -\- c = y the variable x occurs only in 
the term a:^, we get the same value of y for positive and negative 
values of x. Hence the loop which forms the graph of the equa- 
tion is symmetrical with respect to the Y axis. 

114. Degeneration of the quadratic equation. The equation 

ax^ -\-hx -\- c =■ 
has the roots 





_ J 4_ V52 - 4 ac 


Xi — 


2a 


Xo = 





We wish to find the effect on the roots x^ and x^ when a 
becomes very small. If we let a approach 0, then x^ approaches 

an expression of the form -? which must always be avoided. 

Rationalize the numerators and we get 

4 ac 2 c 



2a{- b - ■\/b^ - 4 ac) -b-\^b^-4: ac 
4 ac 2 c 



2a{-b + -^b^-4:ac) - b -{- ^b^ -4.ac 
As a approaches 0, evidently h^ — ^ac approaches J^, Xi ap- 
proaches — j> and x^j since its denominator becomes very small, 

• When — Is positive this involves real factors. If — is negative the factors are 
a a 

imaginary (§ 152). 



GRAPHICAL REPRESENTATION 



105 



increases without limit, that is, approaches infinity. Thus the 
quadratic equation approaches a linear equation when a approaches 
0, and one of its roots disappears since it has increased in value 
beyond any finite limit. The loop-shaped graph of the quadratic 
equation must then approach a straight line as a limit when a 
approaches 0. This is made clear from the following figure, where 
a has the successive values 1, \, y^^, ^V, 0. 

In the figure the curves represent the following equations : 



x-'-l-2 = y.. 


(I) 


x" X 
5-2-2 = ^- 


(H) 


10-2-2 = 2'- 


(III) 


X^ X ^ 

50-2-2 = ^- 


(IV) 


-f-2- 


(V) 



%%-A "■ 


Ml II 171 


^^^J 


7 1 


tsX^ 


t J~ 


>^ \^v ^ 


t t h- 


^§^^A 


l f - ^ 


^SSv^ 


t ^ y^^^ 


^^> 


^ ^'^,^ 


< 


N^^ 




^^^>^ 




^^^-L*- 




^-.K 



In a similar manner we can show that when in the equation 
Jx + c = 0, & approaches as a limit, the root of the linear equa- 
tion becomes infinite. 



EXERCISES 

1. "What real values must k approach as a limit in order that one root of 
each of the following equations may become infinite ? 



(a) Arx^ + Gx + l = 0. 

(c) (A;x - 1)2 - (X + 2)2 = (A: + a:)2. 

(e) V2A:x-H-\/6F=^=V^TT. 

X 



(b) (^2 + 1)3524. a; + 1 = 0. 
(d) A;2 + 4fc2a;2_(aj_i)2 4.2 



(g) 



1 fc + 1 X 
1 X + 1 A;2 



0. 



(f ) Vx-fc + Vx + A; = yfkx + 1. 
1~\ fc2 



/ fcx-1 
'a:x + 1 



(i) 



(^ 



1)^ ^ (fc 



(A; + 1) k 



(X + 1)2 
(j) (A:2 - l)x2 + (A: - l)x + A:2 + 4A; - 5 



106 QUADRATICS AND BEYOND 

2. What real values must k and m approach as a limit in order that both 
roots of the following may become infinite ? 

(a) A:x2 + mx + 1 = 0. 

(b) (2fc-m)x2 + A:x-2 = 0. 

(c) 2 tec2 + (3m - 1 + h)x = 8x2 - 1. 

(d) ijc - l)x2 + (A; + m + l)x + 3 = 0. 

(e) x2 - X - 2(fc + m)a; = (fc + m) (x2 - 1). 

(f) (fc + m)x2 + 2 (fc + m) + 1 = x2 - 2 x. 

(g) (fc + m + l)x2 + (2 A: - m - l)x + 1 = 0. 
(h) (2A; + m + 2)x2 + (4A: + 2m + 3)x + 3 = 0. 

115. Sum and difference of roots. Let x^ and x^ be the roots of 

x'' + bx + G = 0. (1) 

Then (§ 95) x — x^ and x — x^ are factors, and their product 
^'^ — (^1 + ^2)^ + ^1^2 is exactly the left-hand member of (1). 
Consequently the equation 

x^ -\- bx -\- c = x^ — (xi -{- Xz) X + X1X2 
is true for all values of x. Hence by § 96 

- (^1 4- X,) = b, (2) 

X1X2 = c. (3) 

We may state these facts in the 

Theorem. The coefficient of x in the equation s(^-^hx-\-c=0* 
is equal to the sum of its roots with their signs changed. 
The constant term is equal to the product of the roots. 

EXERCISES 

1. Prove the statement just made from the expression for the roots in 
terms of the coefficients (§ 89). 

2. Form the equations whose roots are the following : 

(a) 6, 1. (b) i, I (c) I, 3. (d) - i - 6. 

(e) h h (f) -h+h (g) 2 + V3, 2 - V3. (h) - V3, V3. 

* We should for the present exclude the case where 6^ - 4c<0, since the roots ar, and 
a?j are then imaginary and we have not as yet dettned what we mean by the sum or 
the product of imaginary numbers. We shall see later that the theorem is also true in 
this case. 



GRAPHICAL REPRESENTATION 107 

3. If 4 is one root of ic^ — 3 x + c = 0, what value must c have ? 
Solution : Let Xi be the remaining root. 

Then by (1) _ (xj + 4) = - 3, 

or xi = — 1. 

By (2) c = Xi.4=:(-l)4 = -4. 

4. Find the value of the literal coefficients in the following equations. 

(a) x^-\-bx-9 = 0. One root is 3. 

(b) 0:2 + 4 X + c = 0. One root is 2. 

(c) ax2 + 3x-4 = 0. One root is 2. 

(d) ax2 + 3x + 4 = 0. One root is 7. 

(e) ax2 + 2 X + 6 = 0. One root is 6. 

(f ) x2 + 6x + 4 = 0. One root is - 1. 

(g) x2 - 6x - 6 = 0. One root is - 3. 
(h) x2 + 6x + 6 = 0. One root is - 6. 

(i) 2 x2 — 6 X — c = 0. One root is — 4. 

(j) x2 — 6 X + c = 0. One root is double the other. 

(k) x2 + c = 0. The difference between the roots is 8. 

(1) x2 — 5 X + c = 0. One root exceeds the other by 3. 

(m) x2 — 7 X + c = 0. The difference between the roots is 6. 

(n) x2 — C X + c = 0. The difference between the roots is 4. 

(o) x2 — 3 X + c = 0. The difference between the roots is 2. 

(p) x2 — 2 X + c = 0. The difference between the roots is 8. 

116. Variation in sign of a quadratic. It is often necessary to know the 

sign of the expression 

ax2 + 6x + c 

for certain real values of x, and to determine the limits between which x may 
vary while the expression preserves the same sign. We assume as usual that 
a is positive. 

Theorem L* If the discriminant of ax^ + hx + cis positive, the quadratic 
is negative for all values of x between the values of the roots of the equation. For 
other values of x {excepting the roots) the quadratic is positive. 

* If a were negative, Theorem I would read as follows : If the discriminant is posi- 
tive, the quadratic is positive for all values of x between the values of the roots of the 
equation. For other values of x {excepting the roots) the quadratic is negative. 

When a is negative Theorems II and III may be modified in an analogous manner. 



108 



QUADKATICS AND BEYOND 




\ 



In § 98 we found that when the discriminant of a quadratic equation is 
positive the equation has two real roots. If two roots are real, the loop of 
the graph of the equation ax^ + bx + cz=y cuts the X axis in two points 

(§ 110) as in the figure. The roots are 
represented by A and B, and any real 
value of X between the roots is repre- 
sented by a point P in the line J.JB. 
Since the curve is below the X axis at 
any such point, the value of y, i.e. of 
the expression ax^ + &x + c for values 
of X between the roots, is negative. 

The value of the expression for any 
value of X greater or less than both 
roots is seen to be positive, since for 
such points, for example Q and R, the graph is above the X axis. 

Theorem II. If the discriminant of ax^ -\- hz + c is negative, the expression 
positive for all real values of x. 

When the discriminant is negative the entire graph of ax^ + &x + c = y is 
above the X axis (§ 111), and consequently for any real value of x the corre- 
sponding value of y, i.e. the value of ax^ + &x + c, is positive. 

Theorem III. If the discriminant of ax^ + bx + c is zero, the value of the 
expression is positive for all values of x except the roots of the equation 
ax^ -\- bx + c = 0. 

Hint. See example 16, p. 102. 

We may restate these three theorems and prove them algebraically as 
follows : 

Theorem IV. If the discriminant of the quadratic ax^ + bx-\- cis positive, 
the values of the quadratic and a differ in sign for all values ofx lying between 
the roots, and agree for other values. 

If the discriminant is zero or negative, the value of the quadraiic always 
agrees with a in sign. 

Case I. Since the discriminant is positive, the equation ax^+bx+c=(i 
has two unequal real roots, as Xi and x^, of which we will assume Xi is the 
greater, and we may write the quadratic in the form 

ax2 + 6x + c = a (x — Xi) (x — X2). 

Now for any value of x between Xi and X2 the factor x — Xi is negative, 
while X — Xa is positive, which shows that the quadratic is opposite in sign 
to a for such values of x. For other values of x both these factors are either 
positive or negative, and for such values the quadratic is of the same sign 
as a. 



GKArillCAL KEPRESENTATION 109 

Case II. Since the discriminant 6^ _ 4 ^c is negative and the roots are 

* *!, * & ± V&2 - 4 ac .^ ^^ , -: . 

of the form — — — — - , we may write the quadratic 

2a 

„ , , , / & V62 - 4 ac V , 6 . V62 - 4 ac\ 

V 2a 2a /\ 2a 2a / 

Now for any value of x the expression lx-\ ) is positive, and since 

- \ 2 a/ 
62 — 4ac is negative, 4ac — 6^ is positive; and we observe that tlie last 
member of the equation has the same sign as a. 

Case III. Since the discriminant is zero, the roots are equal and the 
expression has the form 

ax2 4- &x + c = a (x — Xi)2, 

which has evidently the same sign as a, for any value of x. 

EXERCISES 



1. Between what values of x is the expression Vx^ — 5 x + 4 imaginary ? 

Solution : The roots of x^ — 5 x + 4 = are 4 and 1. 

The discriminant A = 62 _ 4 ^c = 25 — 16 = 9 is positive. 
Thus by Theorem I or IV, if 1 < x < 4 f the expression under the radical 
sign is negative and the whole expression is imaginary. 

2. For what values of k are the roots of 

(A:+ 3)x2 + A:x + l = (1) 

(a) real and unequal ? (b) imaginary ? 

Solution : a = k + S, b = k, c = 1. 

A = b^-4ac = k^- 4{k + S) = k'^-ik- 12. 

(a) If A > 0, the roots of (1) are real and unequal. 
The roots of A;^ - 4 A: - 12 are A: = - 2 and 6 
Then, by Theorem II, if 

A; < - 2 or fc > 6, A > 0. 

(b) By Theorem I, if - 2 < A; < 6, A < 0, 
and the roots of (1) are imaginary. 

* See 5 162. t Road " 1 is le6S than x wliich is lew than 4 " or " a; is between 1 and 4." 



110 QUADRATICS AND BEYOND 

3. Determine for what values of x the following expressions are negative, 
(a) aj2 + 2x - 1/ (b) x^-Sx-h 4. 

(c) x2 - 11 oj + 10. (d) x^-15x-{- 60. 

(e)-x2-2x + l. {i) - x^ -\- 7 X + SO. 

4. Determine for what values of k the roots of the following equations 
are (a) real and unequal, (b) imaginary. 

(a) 3 A;x2 - 4 X - 2 = 0. (b) aj2 + 4 fcx + A:2 + 1 = 0. 

(c) x^-{-9kx + 6k + l = 0. (d) x2 + {3k + l)x + 1 = 0. 

(e) (A;2 + 3)x2 + A;x - 4 = 0. (f) 2 x2 - 4x - 2 A; + 3 = 0. 

(g) fcx2 + (4A; + l)x + 4 A: - 3 = 0. (h) {k - l)x2 + 6kx -{- 6k + i = 0. 
(i) {k - l)x2 4- {2k+l)x + A: + 3 = 0. 



CHAPTEE X 

SIMULTANEOUS QUADRATIC EQUATIONS IN TWO 
VARIABLES 

117. Solution of simultaneous quadratics. A single equation 
in two variables, as x^ -\- y^ = 5, is satisfied by many pairs of 
values, as (1, 2), (Vl> V|)> (2, 1), and so on, though there are 
at the same time numberless pairs of values that do not satisfy 
it, as (0, 1), (1, 1), (2, 3). Thus the condition that (x, y) satisfy a 
single quadratic equation imposes a considerable restriction on 
the values that x and y may assume. If we further restrict the 
value of the pair of numbers (cc, y) so that they also satisfy a 
second equation, the number of solutions is still further limited. 
The problem of solving two simultaneous equations consists in 
finding the pairs of numbers that satisfy them both. 

118. Solution by substitution. In this method of solution the 
restriction imposed on (x, y) by one equation is imposed on the 
variables in the other equation by substitution. 

Example. Solve 2x^-\-y'^-l, (1) 

x-y = \. (2) 

Solution : Equation (2) states that x = 1 + y. Thus our desired solution is 
such a pair of numbers that (1) is satisfied and at the same time x is equal 
to y + 1. 

If we substitute in (1) 1 + y for x, we are imposing on its solution the 
restriction implied by (2). 

Thus 2(l + 2/)2 + 2/2=,i^ 

or \ 3 2/2 4- 4 y + 1 = 0. 

The roots are y = — 1, y = — i. 

Corresponding to ?/ = — 1 we get from (2) x = 0. 
Corresponding to y =— \ we get from (2) x = |. 
Thus the solutions are (0, — 1) and (|, — \). 

Ill 



112 QUADRATICS AND BEYOND 

EXERCISES 

Solve the following : 



xy = 4. 


2. 


x-y = 6, 
xy = 36. 


■ x2 + 2/2 = bxy. 


xy — x = 0. 


5. 


x-hy = a, 

x2 + 2/2 =, 5. 


x2 4-2/2 = 50, 
• 9x + 7 2/ = 70. 


'• x-Sy = 0. 


8. 


2x-3y = 4, 
x2 - y2 = 0. 


9 xy = 12, 

2x + 3?/ = 18. 


10 x:y = 9:4, 
^"- a;:12 = 12:y. 


11. 


X2:y2=:a2:62^ 
a — X = b — y. 


12 5x2 + 2/ = 3xy 
2x-2/ = 0. 


*"• 2x-32/ = 0. 




14 (a^ + y)C 


z-2y) = T, 
3. 


-g 3x2-4y = 6x- 
3x+42/ = 10. 


2y% 


16 x2 + ,= 
x:y = 2 


2/2 + X - 18, 
:3. 



^^ ax-by = cy, ^g x2 + 2 X2/ + 2/2 = 7 (x - y), 

a2x2 — 62y2 _ acx2/ + to2. ' 2 X — 2/ = 5. 

^g ax2 + (a-6)x2/-&2/=^=c2, ^q 2x2-5x2/+2/2+10x+12y=100, 

■ (x + 2/) : (X - 2/) = a : 6. 

2j 7(x + 5)2 -9(2/ + 4)2 = 118, 
* X - 2/ = 1. 



x2 + 2/2 = 130, 
23. X + y ^ 





x-y 




2X-2/ + 1 8 


25. 


x-22/ + l~3' 




x2 - 3 X2/ + 2/2 = 5. 


27. 


X2/ + 72 = 6(2x + 2/), 
X 2 




V 8 




4X+2/-1 4X+2/-12 


29. 


2X+2/-1 2X+2/-12 




3x + y = 13. 


31. 


10 9' 
x + 2 + 2/-l = '' 
2 4 



= 2f, 





2x-32/ = l. 


22. 


x2 + 2/2 = a2, 
X m 




2/ n 


24. 


x2 + 2/ + 1 3 
2/2 + X + 1 2 




x — y = l. 


26. 


1 + X + X2^3 

1 + 2/ + 2/^ 
a; + 2/ = 6. • 


28. 


2/ + 6 X 




X — 2/ = 4. 


30. 


«(x-2/)-52/ = 6 
x-2/ 


32. 


? + ? = 3, 

x 2/ 



^^Tri = ;- 5(2/-l) = 2(x + l). 



SIMULTANEOUS QUADRATIC EQUATIONS 113 

119. Number of solutions. We have proved (p. 42) that two 
linear equations have in general one and only one solution. 

Theorem. A quadratic equation and a linear equation ham 
in general two and only two sohUions. 

If the linear equation is solved for one variable, say x, and this 
is substituted in the quadratic equation, we get a quadratic equa- 
tion to determine all possible values of the other variable (i.e. y), 
which must in general be two in number (§ 98). To each one of 
these values of y will correspond one and only one value of x, 
thus affording two solutions of the pair of equations. 

EXERCISES 

1. When may, as a special case, a quadratic and a linear equation 
have only one solution? 

2. When may a quadratic and a linear equation have imaginary 
solutions ? 

3. rind the real values of k for which the following equations have 
(1) only one solution, (2) imaginary solutions. 

x2 + y2 = 16, (1) 

^^^ x-y = k, (2) 

Solution : x = y + k. 

Substitute in (1), {y + fc)2 -{- y"^ = 16. 
or 2 y2 4- 2 A:y + A;2 - 16 = 0. 

As in § 98, a = 2; b = 2k; c = k^ -16. 

Hence A = b'^ - Aac = ^k^ - Sk^ + 12S = - Ak^ + 128. • 

(1) A = when 4k^ = 128, 

or k = ± 4 -^2. There is then only one solution. 

(2) A < when k^ > 32. The solution is then imaginary. 

^^ 2x-y = k. 

x^-y^ = 9, 
^^^ x-2y = k. 

2x2 + 3y2 = 6, 
^J^ x-ky = \. 





x + ky = b, 

^""^ (C2 + 2/2 = 5. 




^ y-x = k, 


^ ^ x^y = k. 


x2 + 2/2 = 25, 
(') 4x-3y = k. 



114 QUADRATICS AND BEYOND 



1 



. 120. Solution when neither equation is linear. In the exam- 
ples previously given one equation has been linear and the other 
quadratic in one or both variables. Often when neither of the 
original equations is linear a pair of equivalent (p. 41) equations 
one or both of which are linear may be found. These latter equa- 
tions may be solved by substitution. 



EXERCISES 

Solve the following equations. 

When neither equation is linear, we can often obtain by addition an 
equation from which by the extraction of the square root a linear equation 
may be found. 

1. x2 + 2/2 = 17, (1) 

xy = 4. (2) 

Solution: x^ + y^ = 17 (3) 

Multiply (2) by 2, 2xy = 8 (4) 

Add x2 + 2 icy + 2/2 ^ 25 

Extract the square root, a; + y = ± 5. (5) 

Subtract (4) from (3), x2 - 2 xy + 2/2 = 9. 

Extract the square root, x — y =±S. (6) 

Solve (5) and (6) as simultaneous equations, 

X + y = ±6, 

x~y = ±S. 

x = +i, +1, -1, -4. 
2/ = +l, +4, -4, -1. 
Thus the solutions are four in number, (4, 1), (1, 4), (—1, — 4), (— 4, —1). 

The following exercise affords another case where a linear equation may 
be found by addition and extraction of the square root. 

2. x^-hxy = 6, (1) 
xy + y^ = 10. (2) 

Solution : Add (1) and (2), 

x^-^2xy + y^ = 16. 
Extract the square root, « + 2/ = ± 4. 

Substitute in (1), x2 + x ( ± 4 - x) = 6, 
x2±4x-x2=:6, 

X = ± f = ± 1^. 

Substitute in (4), 2/ = 2^, — 2|. 

Thus our solutions are (- |, - 2|), (f, + 2J). 



SIMULTANEOUS QUADRATIC EQUATIONS 115 

When neither of the original equations is quadratic, we can often find by 
division an equivalent pair of equations one of which is linear and the other 
quadratic, as in the following exercise. 

3. x3 + 2/8=12, (1) 

X + 2/ = 2. (2) 

Solution: Divide (1) by (2), 

x2 - ajy + y2 ==: 6. (3) 

Square (2), x^ + 2 xy + y^ = 4 

Subtract, — 3 xy = 2 

xy=-l (4) 

Solve (4) with (2) by substitution. 

When the sum of the exponents of the variables is the same in every 
term, the equation is called homogeneous. 

Thus, z'^ + xy^O, 2x^y -3zy^ - 'iz^ - 3y^ = 0. 

When one equation is homogeneous and the other either linear or quadratic 
we may solve them as follows : 

4. 6x2-7xy + 2y2 = o, (1) 

x2 - y = 4. (2) 

Solution : Divide (1) by y^, 

Let- = 2,« 622_7z + 2 = 0. 

y 

Solve for z, z = | or |. 

Thus ?=:ior? = ?. 

y 2 y 3 

Solve (2) with 2 x = y and 3 x = 2 y. 

When both equations are homogeneous except for a constant term we may 
solve as follows : 

5. x2-x2/ + 2y2 = 4, (1) 
2x2-3x!/-2?/2 = 6. (2) 

Solution : Eliminate the constant term by multiplying (1) by 3 and (2) by 2, 
3x2-3x2/+ 6?/2 = i2, (3) 

4 x2 - 6 xy - 4 y2 = 12 (4) 

Subtract (3) from (4), x2 - 3 xy - 10 y2 = o 

* "We observe that y ^0. For if y = were a value that satisfies equation (1), x = 
would correspond. But (0, 0) does not satisfy (2); thus y = Ois not a value that can occur 
in the solutions of the equations. 



116 QUADRATICS AND BEYOND 



X 

Divide by y^ and let - = 2, where y ^d, 

y 

22 _ 32 _ 10 = 0. (S)' 

Factor, (z - 6) (2; + 2) = 0. 

The roots are _ = 6, - = - 2. (6) 

y y 
Solve (6) with (1). 

"When one equation is quadratic in a binomial expression we may solve as 
follows : 



6. x-y 


-Vx-2/ = 2, (1) 




x^-ys = 2044. (2) 


Solution: Let 


Vx - ?/ = z. 


Then (1) becomes 


z2 - z = 2. 


Solving for z. 


z = 2 or - 1. 


Thus 
or 

Solve (3) with (2) as in exercise 


x-2/ = n (3) 
x-y = lj ^ ' 

3. 


7. x^ + y^ = xy = X -\- y. 


8. X3 + 2/3 =7x2/ = 28(x + 2/). 


2/3 + x^y = 4. 


x2^ = a, 
^"- X2/2 = b. 


- x{y-l) = 10, 
' y{x-l) = 12. 


12 a^2 + 2/2-a, 
' xy = b. 


^^ x^y + xy^ = a, 

x^y — xy^ - b. 


.. x + X2/=:35, 
• 2/ + X2/ = 32. 


-g x{x^-{-y^) = 7, 
y{x^-\-y^) = l. 


2x2-32/2 = 6, 
• 3x2-2 2/2 = 19. 


3x2-22/2 = 16. 


-J. 5x2 + 2^2 = 22, 
3x2-52/2 = 7. 


IQ ic2 + ccy + 2/2 = 2, 
• x2 - X2/ + 2/2 = 6. 


20 x + X2/ + 2/ = 5, 

■ a;2 + X2/ + 2/2 = 7. 



Hint. Eliminate x2 or 1/2 as if _ . . o « 

the equations were linear equa- 21. "^ » ^ i'j 

tions in x2 and y^. X2/ = 2 x - 2/ + 9. 

22 «2_x2/ + 2/2 = 37, (x + 2/)(8-x) = 10, 

• x2 - j/2 = 40. ■ (X + 2/) (5 - 2/) = 20. 

24 (»2 + 2/2)(x + 2/)=6, 25 (^ + 2/)^ = 3x2 - 2, 

' «y(x + 2/) = - 2. * • (X - 2/)2 = 32/2 - 11. 



SIMULTANEOUS QUADRATIC EQUATIONS 



117 



26. 


X + y/x^y = a, 
y + ^x?/2 = 6. 


28. 


X + y = 58, 


Vx + Vy = 10. 


30. 


X + 2/ = 3. 


32. 


4x*-9y2 = o, 

^ —O ■ ..O O / 1 -A 



34. 



38. 



40. 



42. 



4x2 + 2/2 = 8(x + y). 
3x2/-2(x + y) = 28, 
2x2/-3(x + 2/) = 2. 

x2 + ?/2 + X + y = 18, 
x2 - 2/2 + X - y = 6. 

3x2- 2 2/2 = 6(x-2/), 
X2/ = 0. 

x2-X2/ + 2/2 = 13(x-2/), 
xy = 12. 



Vx(l-y) +V2/(l-ic) = a, 



2/ 



^^ Vl-x2Vl-2/2 + X2/ = |l, 
• x-y = ll. 



X _y _ 16 
46. y X " 16' 

3x2 + 6y2 = i20. 

X V^ + 2/ Vy _ 1 
48. xy/x — yy/y 2 



x3 - 8 = 8 - y3. 



V2/ — Va — X = Vy -X, 
50. Vy — X + Va — x _ 5 



52. 



Va — X 



1 , 1_3 

i + i = l 
x2 y« 4 



27. 

29. 
31. 

33. 
35. 
37. 
39. 
41. 

43. 
45. 

47. 
49. 

51. 
53. 



xVx + y = 3, 
y Vx + y = 1. 

■y/x +y/y = a, 
X + y = 6. 

Vx — Vy = 2, 
(x + y) Vxy = 610. 



Vx-6 + Vy + 2 = 5, 
X + y = 16. 
xy + xy-i = x2 + y2, 
xy-xy-i = 2(x2 + y2). 

x-2 + 2 y-2 = 12, 

x-2 - x-iy-i + y-2 = 4. 

5c2 + y2_5(a; + y)^8, 
x2 + y2-3(x + y) = 28. 

2x2-3xy + 6y- 6 = 0, 
(x-2)(y-l) = 0. 



V6-3x + x2 + V6-3y + y2 = 6, 
X + y = 3. 



X y ' 



y = .3. 



a y _ 25 

y '^x~12* 
x2 - y2 = 28. 



V 5x /x + 

a; + y \ 6a 
icy - (x + y) = 1. 

^ + ?^ = 2, 
a2 62 

bx + ay _m 

bx — ay n 

x8 _ y3 _ 16 
y X 2 ' 

- -1=-. 
y x ~ 2' 



y ^ 3 V2 
X 2 ' 



Hint. Let - = u, - = v. 

a ' y 



118 QUADRATICS AND BEYOND 

X -1 a—1 






2, 
54. ) ^i 55. 

3. 



y-\ b- 
x3 - 1 a3 



yS _l 53 _ 1 

lift ^ + ^ 57 ^^^ 

^ 2 x + l 221 

1 1_5 1_?-1 

5g x'^y~6' 59 "^ ^~ ' 



X + 22/ + I2 x + 

xy = 2. 

PROBLEMS 

1. Two numbers are in the ratio 5 : 3. Their product is 735. What are 
the numbers ? 

2. Divide the number 100 into two parts such that the sum of their 

squares is 5882. 

3. The sum of the squares of two numbers increased by the first is 
205 ; if increased by the second the result is 200. What are the numbere ? 

4. The diagonal of a rectangle is 85 feet long. If each side were longer 
by 2 feet, the area would be increased by 230 square feet. Find the length 
of the sides. 

5. The diagonal of a rectangle is 89 feet long. If each side of the rec- 
tangle were 3 feet shorter the diagonal would be 85 feet long. How long 
are the sides ? 

6. The sum of two numbers is 30. If one decreases the first by 3 and 
the second by 2 the sum of the reciprocals of the diminished numbers is ^. 
What are the numbers ? 

7. The sum of the squares of two numbers is 370. If the first were 
increased by 1 and the second by 3, the sum of the squares would be 500. 
What are the numbers? 

8. A number of persons stop at an inn, and the bill for the entire party 
is $24. If there had been 3 more in the party, the bill would have been 
$33. How many were in the party and how much did each pay ? 



SIMULTANEOUS QUADRATIC EQUATIONS 119 

9. A fruit seller gets $2 for his stock of oranges. If his stock had con- 
tained 20 more and he had charged f of a cent more for each, he would have 
received $3 for his stock. How many oranges had he and how much did he 
get apiece for them ? 

10. A man has a rectangular plot of ground whose area is 1250 square 
feet. Its length is twice its breadth. He wishes to divide the plot into a 
rectangular flower bed, surrounded by a path of uniform breadth, so that 
the bed and the path may have equal areas. Find the width of the path. 

11. In going 7500 yards one of the front wheels of a carriage makes 1000 
more revolutions than one of the rear wheels. If the wheels were each a yard 
greater in circumference, the front wheel would make 625 more revolutions 
than the rear wheel. What is the circumference of the wheels ? 

12. A man has |539 to spend for sheep. He wishes to keep 14 of the 
flock that he buys, but to sell the remainder at a gain of $2 per head. 
This he does and gains $28. How many sheep did he buy and at what 
price each? 

13. A man buys two kinds of cloth, brown and black. The brown costs 
25 cents a yard less than the black, and he gets 2 yards less of it. He 
spends $28 for the black cloth and $25 for the brown. How much was each 
a yard and how many yards of each did he get ? 

14. A man left an estate of $54,000 to be divided among 8 persons, namely, 
his sons and his nephews. His children together receive twice as much as 
his nephews, and each one of his children receives $8400 more than each one 
of his nephews. How many sons and how many nephews were there ? 

15. A and B buy cloth. B gives $9 more for 60 yards than A does for 
45 yards ; also B gets one yard more for $9 than A ddes. How much does 
each pay? 

16. A sum of money and its interest amount to $22,781 at the end of a 
year. If the sum had been greater by $200 and the interest :^ of 1 per 
cent higher, the amount at the end of the year would have been $23,045. 
What was the sum of money and what was the interest ? 

17. If one divides a number with two digits by the product of its digits, 
the result is 3. Invert the order of the digits and the resulting number is in 
the ratio 7 : 4 to the original number. What is the number ? 

18. What number of two digits is 4 less than the sum of the squares of 
its digits and 5 greater than twice their product ? 

19. Increase the numerator of a fraction by 6 and diminish the denomi- 
nator by 2, and the new fraction is twice as great as the original fraction. 
Increase the numerator by 3 and decrease the denominator by the same, and 
the fraction goes into its reciprocal. What is the fraction ? 



120 QUADRATICS AND BEYOND 

121. Equivalence of pairs of equations. In the theorems of 
this section the capital letters represent polynomials in x and y, 
and the small letters represent numbers not equal to zero. 



I 



Theorem I. The pairs of equations 
A 

are equivalent. 

If (ari, 2/1) be a pair of values that satisfy (1), then when x and 
y in 5^ are replaced by Xi and yi the equation B"^ = i^ is a numer- 
ical identity. These values (xy, y^ must then satisfy one of the 
equations -S = ± ^, for if they did not, but only satisfied the equa- 
tion say B = c when c =^ ±b, then the hypothesis that B^ = }p- is 
satisfied by (cci, ?/i) would be contradicted. 

Conversely^ any pair of values that satisfy B = -^h evidently 
satisfy B"- = b\ 

This theorem is used, for instance, in exercise 2, p. 114, and justifies the 
assumption that 

are equivalent pairs of equations. 

Theorem II. The pairs of equations 

are equivalent. 

li A = a and B = h are satisfied by a pair of numbers (xj, y^), 
we multiply the identities and obtain AB = ab. 

Conversely f if A = a, AB = ab are identically satisfied by a 
pair (xi, yi), since a ^ we can divide the second identity by 
the first and obtain B = b. Thus if (cci, yi) satisfy one pair of 
equations they satisfy the other pair. 

This theorem is assumed in exercise 3, p. 115, to show that 

a;« + w« = 12 ^ , a;2 - XM + 7/2 = 6 1 . , ^ 

, ^ y and . " ~r ^ ^ are equivalent. 



SIMULTANEOUS QUADRATIC EQUATIONS 121 



Theorem III. The pairs of equations 



^:^}a) ^f!^!:^K^) 



B = 0]^ ' cA-\-dB = 
are equivalent where a, b, c, and d are numbers such that 

ad — be ^ 0. 

If (xi, yi) satisfy (1), evidently they also satisfy (2). Thus all 
solutions of (1) are among those of (2). 
Conversely J if (xi, ^/i) satisfy (2), then 

__bB__dB 
a c 

Thus (ad — hc)B = 0. 

Thus since (ad — bo) =^ 0, . 

^ = 0. 

Similarly, ^ = 0. 

This theorem has been assumed in exercises 1, 2, 3, 6, p. 114. In 1, for example, 
it is necessary to show that 

are equivalent. In this case a=c=l, b = — d = 2. Thus ad — be = — 4: ^t 0. 

122. Incompatible equations. When a pair of simultaneous 
equations can be proven equivalent to a pair of equations which 
contradict each other or are absurd, they are incompatible and 
have no finite solution. 



(1) 
(2) 



Example 1. 


xy = 1 


Subtract, 


xy = -1 
0= 2 


Example 2. 


x'^ + y^ = 4, 
4x2 + 42/2 = 49. 


Multiply (1) by 4, 
Subtract, 


4x2 + 42/2 = 16 

4x2 + 42/2 = 49 

= 33 



122 



QUADRATICS AND BEYOND 



123. Graphical representation of simultaneous quadratic equa- 
tions. Every equation that we have considered may be rep- 
resented graphically by plotting 
in accordance with the method 
already given (p. 93). 

The solution of simultaneous 
equations is represented by the 
points of intersection of the cor- 
responding graphs. 

Thus the equations 

x'^,f = 25, 
2xy = 9 
have the solutions 




a;=±2± 



V34 



y==F2± 



V34 



or 



X = 4.9, .9, - .9, - 4.9, 



y 



.9, 4.9, - 4.9, - .9. 



These equations have as their graph the preceding jfigure. 
The equations 



ic2 + 2£cH-4?/ + l = 0, 
x-f22/ + 4 = 0, 

which have the solutions 

a; = ± Vt = ± 2.6, 

y = - 2 =F V^ = - 3.3 or - 



Y'^ 



- ^ - 




^^ 


^N, X 


Z '^ 


^J\ - 


/ 


^\r - 


r 


%> 


7 


L^ 


!-—- 


5: 



r> 


~fcM — 


^^ ^ 


, Ny 


1 /T 


"\ V 


I ^ 


^ I^ 


-\ ^ ° 


-4 t* 


-^^^- 


J t 


_ ^^ 


7 




"^ 



have as their graph the figure shown 
above. 

As in the case of linear equations, 
incompatible equations afford graphs 
which do not intersect. Thus the graph 
of the equations in example 2, p. 121, is 
found to be two concentric circles, as 
is shown in the adjacent figure. 



SIMULTANEOUS QUADRATIC EQUATIONS 



123 



Simultaneous equations which have 
imaginary solutions also lead to non-inter- 
secting graphs (p. 101). 

Thus the equations 

a;2 + 2/' = 4, 
have the adjacent figure as their graph. 



. .^.- 


. 


^, 






V : 


■ /"=VlXI 1 1 




1 >L f 





J hs' 


^^^/| 1 1 U" 







EXERCISES 

1. Inteipret the graphical meaning of equivalent pairs of equations. 

2. Plot and solve x^ + ?/2 = 2, 

X + 2/ = 2. 
What general statement concerning the graphical meaning of a single 
solution of quadratic and linear equations does this example suggest ? 

3. Plot and solve the following : 



(a) 



25. 



X2 + 2/2 

4x2 + 9^2 = 144. 



(b) 



(c) 



x2 + 2/2 = 25, 



x2 + y2 = 25, 

4x2 -8x + 92/2 = 140. 



5x2 + 2/2 = 25. 

What general statement concerning the graphical interpretation of four, 
three, or two real solutions of equations do these examples suggest ? 

4. State the algebraical condition under which two quadratic equations 
have four, three, two, or one real solutions (see p. 113). 

5. Plot and solve the following : 
x2 + 2/ = 0, „ . a;2 + 2/2 = 9. 

32/ = 0. 



(a) 



(c) 



(e) 



X2 



4x-22/ = 3, 



^.2 I 2/2 
y^' x2 - y2 = 0. 



xy 



0. 



^ ' X2 + y2 



16. 



4x2 + 92/2 = 36, 

x2 = — 4 ?/. 



(f) 



xy = 1, 
2x-Sy = lS. 



124. Graphical meaning of homogeneous equations. Consider for example 
the homogeneous equation 

3x2 -10x2/ -82/2 = 0. (1) 



If we let z = - , we get 

X 



or 
or 



3 _ 10 z - 8 z2 = 0, 

8^2 + 102-3 = 0, 

(4z-l)(2z + 3) =0. 



124 



QUADRATICS AND BEYOND 



The roots are 


2 = 1 and 2 = - |. 


Thus 


y=i and y=-l, 

a; 4 X 2 




4y-x = and 3x + 22/ = 0. 





or 

These equations represent two straight lines through the origin which 
taken together form the graph of equation (1). This 
example may obviously be generalized : Any homo- 
geneous equation of the form ox^ + hxy + cy"^ = with 
positive discriminant represents two straight lines 
through the origin. Such an equation is equivalent 
to two linear equations. 

In an example like 5, p. 115, we obtain in place 
of the given pair of equations a pair of equivalent 
equations one of which is homogeneous and the other 
of which is factorable. We can learn the graphical 
meaning of tl;is method of solution by studying a 
particular case. Consider for example the equations : 

jc2 + 2x2/ + 7?/2 = 24, (1) 

2x^-xy-y^ = S. (2) 

By eliminating the constant terms we obtain the product of the two 
equations x + y = and x — 2 y = 0. Thus the problem of solving (1) and 
(2) is replaced by that of solving the two following pairs of equations : 

x2 + 2X2/ + 72/2 = 24, (^) (2) . (2) 

x + y = 0, 
or x2 + 2x2/ + 73/2 = 24, 

X - 2 2/ = 0. 

The graphical meaning of this 
method of solving the equations (1) 
and (2) is seen in the fact that the 
problem of finding the points of inter- 
section of the graph of equation (1) 
with that of (2) is changed to that of 
finding the intersection of the graph of (1) with a pair of straight lines. 
This appears in the figure where the curves and lines are numbered as 
above. The closed curve represents (1). 




CHAPTEE XI 
MATHEMATICAL INDUCTION 

125. General statement. Many theorems are capable of direct 
and simple proof in special cases, while for the general case a 
direct proof is difficult and complicated. 

If we ask whether ic" — 1 is divisible by x — 1, it is easy to 
make the actual division for any particular value of n, as n = 2 
or n = 3. But if x^ — 1 is shown divisible by a; — 1, we are no 
wiser than before concerning the divisibility of x^ — 1. Suppose, 
however, we can prove that the divisibility for ti = m -f 1 follows 
from that for n = m, whatever value m may have. Then since 
we have established the fact by direct division for n = 3, we may 
be assured of the divisibility for ?i = 4, then for n = 5, and so on. 

Now x'^+^ — l=x(x"' — l)-\-(x — l) 

is identically true. If ic — 1 is a factor of cc"* — 1 for a given value 
of m, it is a factor of the right-hand member and consequently a 
factor of the left-hand member (§ 69), which was to be proved. 
Thus the divisibility of x"" — 1 hjx — 1 is established for any 
integral value of n greater than the one for which the division 
has actually been carried out. 

To complete the proof of a theorem by mathematical induction 
we must make two distinct steps. 

^-^^^st Establish the theorem for some particular case or cases, 
preferaUy for n =1 and n = 2. 

Second. Show that the theorem for n = m -{- 1 follows from 
its assumed validity for n = m. 

Example. Prove that the sum of the cubes of the integers 

from 1 to 71 is SK^(^ + 1)]S'- 

To prove that 1« -f- 2» 4- 3* + • • • -F 7i» = J^[7i(n + 1)]^. 

125 



126 QUADRATICS AND BEYOND 

First. This theorem is true for n = 1. 

For 1» = 1 = J ^ [1 (2)] p = 12 = 1 

The theorem is also true for n = 2. 

For l» + 2» = 9= J^[2(2 + l)]P = G-6)2=32 = 9. 

Second. Assume the theorem for n = m,* 

1« + 2« + . . . + w« = J J[m(m + 1)]P. 

Add (m + 1)^ to both sides of the equation, 

1« + 2« + . . • + m« + (m + 1)« = ^[m(m + 1)] p + (m + 1)* 

= [(^m)2-|-m + l](m + l)2 



= ( 4 ) (^ + ly 



= [i(mH-l)(m + 2)]S 
which is the form desired, i.e. m + 1 replaces m in the formula. 

EXERCISES 
Prove by mathematical induction that 

1. 1+3+6 + .. .+(2n-l)=n2. 

2. 2 + 22 + 28 + . . . + 2" = 2(2» - 1). 

3. 3 + 6 + 9 + ... +3n = f n(n + l). 

4. 12 + 22 + 32 + ... + n2 = ^n(n + l)(2n + l). 

5. 13 + 28 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2 

6. 42 + 72 + 102 + . . . + (3 n + 1)2 = ^ n(6 n2 + 16n + 11). 

7. X" — y" is divisible by x — y for any integral values of n. 

8. x2» — 2/2" is divisible by x + y for any integral values of n. 

9. 1.2 + 2-3 + 3. 4 + 4.6 + . .. + n.(n + l) = ^n(n + l) (n + 2). 

10. 1 • 1 + 2 • 32 + 3 . 62 + ... + n(2 n - 1)2 = ^n(n + l)(6n2 - 2n - 1). 

11. 1.2.3+2.3.4+3.4.6 + ...+n(n+l)(n+2) = in(?i+l)(n+2)(n+3). 

12. (1« + 28 + 38 + • . . + n8) + 8(16 + 26 + 35 + . . . + n6) 

= 4(1 + 2 + 3+ ... +n)8. 

* This statement does not imply that we assume the validity of the formxila for any 
values for which it has not yet been established, but only for values of m not greater 
than 2. 



MATHEMATICAL INDUCTION 127 



14. 


Ill 1 _ ^ 

1.2'2.3'3.4' 'n-(w + l) n + 1 


15. 


o 


16. 


2.6 + 3.6 + 4.7 + ... + (« + l)(. + 4) = "<» + ''g><'' + «V 



17. 2.4 + 4.6 + 6.8 + -.. + 2n(2n + 2) = ^(2n + 2)(2n + 4) 

o 

18. A pyramid of shot stands on a triangular base having m shot on a 
side. How many shot are in the pile ? 






CHAPTEE XII 

BINOMIAL THEOREM 

126. Statement of the binomial theorem. When in previous 
problems any power of a binomial has been required we have 
obtained the result by direct multiplication. We can, however, 
deduce a general law known as the binomial theorem, which gives 
the form of development of (a + hy, where n is any positive integer 
and a and h are any algebraical or arithmetical expressions. This 
law is as follows : 

(a 4- ^»)« = a« + ^ a«- ^b + ^ ' ^f ~ ^^ a^-^'b^ + "- + b\ 
From this expression we deduce the following 
EULE FOR THE DEVELOPMENT OF (a + bf. 

The first term is a". 
n 
1 

To obtain any term from the preceding term, decrease the 
exponent of a in the preceding term by 1 and increase the 
exponent of b by 1 for the new exponents. Multiply the coefficient 
of the preceding term by the exponent of a, and divide it by the 
exponent of b increased by 1 for the new coefficient. 

Remark. In practice it is usually more convenient first to write down all the 
terms with their proper exponents, and then form the successive coefficients. 

EXERCISES 

Verify by multiplication the rule given for the following : 

1. (a + 6)». 2. (x - 2/)8. 

3. (2a + 36)*. 4. (v^+Vi^)'. 

5 (2 a -6)4. 6. {x-y/y)\ 

7. (3a -2 6)8. 8. (a-ia; + 6-iy)4. 

128 



BINOMIAL THEOREM 129 

127. Proof of the binomial theorem. We have already stated 
that / -j \ 

{a + ^)« = a" + 7 a--^b + ""^ ~ ^ a^-'b"" + ••. (1) 

and have seen that it is justified for every particular case that 
we have tested. By complete induction we now prove this 
theorem when ti is a positive integer. 

First. Let n = 2. 

That is, (a -{- by= a^ -{- 2 ab + b\ 

This expression evidently obeys the law as stated in (1). 
Second. Assume the theorem for n = m. 

That is, (a + &)- = a- 4- ^ a^-^h + '^^'^ -^) ^m-2^2 + .... (2) 

Multiply both members hj a -\-b, 

(a 4- Z>)"'+i = a'«+i ^ja'^h^j a'^-^b'^ 

z 

Z 
This expression is identical with (2) except that (m + 1) 
replaces m. Hence the theorem is established so far as the 
first three terms are concerned. 

128. General term. Though we have stated the binomial 
theorem for a>§eneral value of ri, we have only established the 
exact form of the first three terms. 

Let {a + by = a'* + c^a^'-^b + Cga"" ^'^ ^ . , 

We note that the sum of the exponents of a and 5 is w in any 
term of the development of {a -\- by. Also the exponent of b in 
the (r -\- l)st term is r. 

We have already seen that 

n n(n — l) 

and that the first three terms are 

J. ± • z 



130 QUADKATICS AND BEYOND 

respectively. This indicates that the coefficient of the next term 

will be — ^^ — - — -^r-^ and in general that the coefficient of the 

(r + l)st term has the form 

_ n(n-l)---(n-r + l) 

l-2...r ' W 

which is in fact the form that our rule (§ 126) would afford for 
any particular value of r. 
This affords the following 

Rule. The (r + l)st term of {a + hf is 

n{n-l)-"{n-r + l) ^„_ , ^, 
1.2'- r 

The form of the coefficient may be easily remembered since the 
denominator consists of the product of the integers from 1 to r, 
while the numerator contains an equal number of factors consist, 
ing of descending integers beginning with n. 

For any particular values of n and r we could easily verify the 
rule by direct multiplication. For the rigorous proof see p. 178. 

EXERCISES 

Develop by the binomial theorem : 



/_a_ _ ^^Y 



Solution 



6-5-4-3 (_±y(_ VxV 6.5.4.3-2 / a V/ VaV 

1.2.3.4VV^/V a2/ l-2.3.4.6VVi/\ aV 
6. 5-4. 3.2.1 / V^y 
1.2.3.4.5.6V a2/ 

x« x2 X a3 a6 a* "^ a^a" 

2. (f-fa)6. 3. (Va + v'ft)'. 



BINOMIAL THEOREM 131 

10. (1 + V^y - (1 - V^)'. 11. ( Vx + v^)*+ ( Vx - Vy)*. 

(2 X 3 v\^^ 
1 ) • 
Sy 2x7 

Solution : n = 10, r + 1 = S. 

The (r + l)st term of (a + 6)" is (§ 128) 

n(n-l)-'-(n-r + 1) 

1-2. -r 
In this case we get 

10-9-8.76-5.4 (^xy/Syy 

1.2.3.4-6-6.7 '\3y)\2x) 

33y3 27x7 24X* 

_ 120-81 y^_ 1215y4 
16 'x*~ 2x* ' 

(1\13 
a + -) . 

14. Find the 6th term of (— - ^V^. 

\2y xj 

15. Find the 8th term of ( — - ^ ) . 

\y X / 

16. Find the 6th term of l2aVb ) . 

V 2aVb/ 

17. Find the 7th term of (^ - ^) • 

18. Find correct to three decimal places (.9)^ 
Solution; (.9)8 = (1 - .1)8 

8^7^6^ ^ 

1.2.3.4^ / V ' -r 
= 1 - 8 • 0.1 + 28 • 0.01 - 66 . 0.001 + 70 • 0.0001 
= 1 + 0.28 + 0.0070 - 0.8 - 0.056 
= 1.2870 - 0.856 = .431. 

In this exercise any terms beyond those taken would not affect the first 
three places in the result. 



132 QUADRATICS AND BEYOND 

Compute the following correct to three places : 

19. (1.1)10. 20. (2.9)8. 21. (.98)11. 22. (1.01)«^ 

23. (1^)8. 24. (1^)10. 25. (98)8. 26. (203)5. 

27. In what term of (a + 6)2o does a term involving ai* occur? 

28. For what kind of exponent may a and h enter the same term with 
equal exponents ? 

29. For what kind of exponent is the number of terms in the binomial 
development even ? 

30. Find the first three and the last three terms in the development of 

(Z, . 1 \24 



CHAPTEB XIII 
ARITHMETICAL PROGRESSION 

129. Definitions. A series of numbers such that each numbei 
minus the preceding one always gives the same positive or 
negative number is called an arithmetical series or arithmetical 
progression (denoted by A.P.). 

The constant difference between any term and the preceding 
term of an A.P. is called the common difference. 

The series 4, 7, 10, 13, • • • is an A.P. with the common difference 3. The series 
8, 62, 5, 32, • • • is an A.P. with the common difference — §. The series 4, 6, 7, 9, 
10, • • • is not an A.P. 

EXERCISES 

Determine whether the following series are in A.P. If so, find the common 
difference. 

1. 6, m If, .... 2. 27, 22^, 18, .... 

3. 6, 4^ 3, H, .... 4. 6, -2, -8, .... 

5. VI, V2, 3 VI, •••. 6. 8, 5|, 3|, If, .... 

- 1 2 4 V2-I V2 1 

'• V2 V2 V2 2 2 2(V2-1) * 

9.3, -^, -3f, -6f, .... 10.^ ^^ + 2 -^^ 



6 6(V3-4) 

130. The nth. term. The terms of an A.P. in which a is the 
first term and d the common difference are as follows: 

a, a -i- d, a -\- 2d, a -{- 3dj '". (1) 

The multiple of d is seen to be 1 in the second term, 2 in the 
third term, and in fact always one less than the number of the 
term. If we call I the nth. term, we have 

I = a -{■ (^n — 1) d. 

133 ^ .., a 




134 QUADRATICS AND BEYOND 

We may also write the series in which I is the nth term as* 

follows : 

a, a-{- dy a + 2d, • • • , I — 2d, I — d, I. 

131. The sum of the series. We may obtain a formula for 
computing the sum of the first n terms of an A.P. by the following 

Theorem. The sum s of the first n terms of the series 
a, a -^dy • • •, I — d, I is 

By definition, 

s = a-\-(a + d) + (a + 2d)-\ -]-{l - 2d)-\-(l - d)+ I. (1) 

Inverting the order of the terms of the right-hand member, 

s =. I + {l - d) + {I - 2 d) -^ ' ■ - + {a + 2 d) + (a -\- d)+ a. (2) 
Adding (1) and (2) term by term, 
2s^{l-\-a)+{l + a)-\-{l + a) + .'-^{l-\-a) + (l-\-a) + {l + a) 

z=n{a-\- 1). 
Thus s = ^(a-\-l). 

132. Arithmetical means. The terms of an A.P. between a 
given term and a subsequent term are called arithmetical means 
between those terms. By the arithmetical mean of two numbers 
is meant the number which is the second term of an arithmetical 
series of which they are the first and third terms. Thus the 

arithmetical mean of two numbers a and h is — r— > since the 

numbers a, — ^r— > h are in A.P. with the common difference — — • 

The two formulas 

« = a + (»i - 1) d, (I) 

« = -(« + 1) (II) 

contain the elements a, I, s, n, d. Evidently when any three are 
known the remaining two may be found by solving the two equa- 
tions (I) and (II). 



AKITHMETICAL PROGRESSION 135 

EXERCISES 

1. Find the 16th term and the sum of the series 4, 2, 0, — 2, • • .. 
Solution: n = 16, a = 4, d = 2 - 4 = - 2. 

Z = a + (n - l)d = 4 -t- 15(- 2) = -26,^ 

« = |(«4-0 = f(4-26)=-176. 

2. Z = 42, a = - 3, d = 3. Find n and s. 

3. a = - 4, n = 8, s = 64. Find d and I. 

4. d = - i, n = 6, « = 21. Find s and a. 

5. d = — i, n = 10, s = 65. Find a and i. 

6. s = 161, Z = 4, a = — 3. Find d and n. 

7. Z = 22, s = 243, n = 13. Find a and d. 

8. s = - 15, Z = - 2, d = 2. Find n and a. 

9. d = 41, a = - 16, s = 140. Find n and i. 

10. Insert 8 arithmetical means between 4 and 28. 

11. Find expressions for n and s in terms of a, I, and d. 

12. Find expressions for I and a in terms of s, n, and d. 

13. Find expressions for a and s in terms of d, /, and n. 

14. Find expressions for d and n in terms of s, a, and Z. 

15. Find the 13th term and the sum of the series 

V2-1 V2 1 

2 ' 2 '2(V2-l)'"" 

16. Find the 10th term and the sum of the series 

V3 3V3 + 2 V3 2 
T"'~~6~"'"2" + 3'"- 

17. Insert 4 arithmetical means between — - and 

V2 2 

10\/6 



18. Insert 6 arithmetical means between -x - and 



4 



19. Insert 3 arithmetical means between and 

2 

20. Find the 21st term and the sum of the series , V^ 

V2 

V2 

21. Find the 10th term and the sum of the series — = 




136 QUADRATICS AND BEYOND 

22. Eind expressions for d and a in terms of s, Z, and n. 

23. Find expressions for d and I in terms of a, n, and s. 

24. Find the 8th term and the sum of the series x, 4ic, 7x, • • •. 

25. Find the 9th term and the sum of the series 8, 9J, 10|, • • •. 

26. Find the 12th term and the sum of the series 8, 7y\, 6|, • • • . 

27. Find the 8th term and the sum of tlie series — 8, — 4, 0, • • •• 

28. Find the 12th term and the sum of the series 27, 22 1, 18, • • •. 

29. Find the 20th term and the sum of the series 1, — 2i, — 6, • • •. 

30. Find the 11th term and the sum of the series 5, — 3, — 11, • • •. 

31. Find the 9th term and the sum of tlie series x — y, x, x -]- y, • • -. * 

32. Insert n — 2 arithmetical means between a and I. Write the first tliree. 

Kemark. Often an exercise may be solved more simply if instead of assum- 
ing the series x, x-\- y, z + 2y, ■■ -we assume x — y, x, x -\- y when three terms 
are required, or x — 2y, x — y, x, x + y, x + 2y when five terms are required, or 
x~-3y,x — y,x + y,x-i-3y when four terms are required. 

33. The sum of the first three terms of an A. P. is 15. The sum of their 
squares is 83. Find the sum of the series to ten terms. 

34. Find expressions for n and a in terms of s, Z, and d. For what real 
values of s, Z, and d does a series with real terms not exist ? 

35. In an A. P. where a is the first term and s is the sum of the first 
n terms, find the expression for the sum of the first m terms. 

36. Find expressions for n and I in terms of a, s, and d. For what real 
values of a, s, and d does a series with real terms not exist ? 

37. If each term of the series (1), § 130, is multiplied by m, is the new 
series in A. P., and if so, what are the elements of the new series ? 

38. If each term of the series (1) in § 130 is increased by 6, is the new 
series in A. P., and if so, what are the elements of the new series ? 

39. The difference between the third and sixth terms of an A. P. is 12. 
The sum of the first 10 terms is 45. Find the elements of the series. 

40. Find the 10th term of an A. P. whose first and sixteenth terms are 3 
and 48. Find also the sum of those eight terms of the series the last of 
which is 60. 

41. Two A.P.'s have the same common difference, and their first terms 
are 2 and 4 respectively. The sum of the first seven terms of one is to the 
sum of the first seven terms of the other as 4 is to 6. Find the elements of 
both series. 

42. The three digits of a number are in A. P. The number itself divided 
by the sum of the digits is 48. The number formed by the same digits in 
reverse order is 396 less than the original number. What is the number? 



CHAPTEE XIV 
GEOMETRICAL PROGRESSION 

133. Definitions. A series of nTimbers such that the quotient 
of any term of the series by the preceding term is always the 
same is called a geometrical progression (denoted by G.P.). 

The constant quotient of any term by the preceding term of a 
G.P. is called the ratio. 

The G.P. series 4, 8, 16, • ■ • has the ratio 2. The G.P. series 8, 4 Vi, 4, • • • has 

V9 

the ratio J-^i . 

^' EXERCISES 

Determine which of the following series are in G.P. and find the ratio. 
1. 4, 2, 1, .... 
3. 8, -2, .5, .... 

5. Vl» i» Vl, •••• 



2. 


4, 8, 16, .... 


4. 


8, -4, -2, .... 


6. 


6,-21, 73i, ... 


8. 


^ 2 ^ 

V2' '' V2' 


10. 


V3 [3" V3 
8 ' V32' 4 ' 



7.-^, A 2, .... 

V2 

\ 6 V5 Vl5 

11. —J^ — -^, 5-2V6,3V3-V2,.... 12. V2 - 1, 1, V2' + 1, .... 

V3 - V2 

134. The nth term. The terms of a G.P. in which a is the 
first term and r the ratio are as follows : 

a, ar, ar^^ at^, .... 

The power of r in the second term is 1, in the^ third term is 
2, and in fact is always one less than the number of the term. 
If we call I the nth term, we have the following expression for 
the T^th term : 

137 ^ 



138 QUADRATICS AND BEYOND 

135. The sum of the series. We obtain a formula for finding 
the sum of the first 7i terms of a G.P. by the following 

Theorem. The sum s of the first n terms of the geometrical 
progression a, ar, ar^^ ... is 

a — rl .{■> 
' = T^r- " 
By definition, s = a -\- ar + ar^ H- • • • 4- ar"~^ 
= a(l -h r + r"^ -\ h r^~^) 



K^) 



a — rar"^ ^ a — rl 



by (III), p. 15 
by (I), p. 137 



136. Geometrical means. The n — 2 terms between the first 
and the ?ith term of a G.P. are called the geometrical means 
between those terms. 

If one geometrical mean is inserted between two numbers, it 
is called the geometrical mean of those numbers. Thus the 
geometrical mean between a and h is ^ ah. 

The two fundamental formulas 

« = ar»*-i, (I) 

_ a{\-r^) _ a-rl 

^ - 1-r - "rr7 ^^^^ 

contain the five elements a, I, r, n, s, any two of which may be 
found if the remaining three are given. 

EXERCISES 

1. Find the 7th term and the sum of the G.P. 1, 4, 16, . . .. 
Solution : a = 1, n = 7, r = 4. 

Substituting in (I), I = ar*'-^ = 1 • 4^ = 4096, 

a — rl 



1-r 



o V *-^ *• • /TTv 1-4-4096 1638.3 ^,^, 
Substitutmg m (II), s = = = 6461. 



GEOMETRICAL PROGRESSION 139 

2. Insert 2 geometrical means between 4 and 32, 

3. Insert 4 geometrical means between 32 and 1. 

4. Insert 3 geometrical means between 3 and |f . 

5. Insert 4 geometrical means between a^ and l^. 

6. Insert 4 geometrical means between 1 and 9V3. 

7. Insert 3 geometrical means between y- a-iid 73|. 

8. What is the geometrical mean between 3 and 27 ? 

9. Insert 3 geometrical means between VS and "v/24. 

10. Insert 4 geometrical means between a and a^ Va6^. 

11. Insert 3 geometrical means between — | and — 2^. 

12. What is the geometrical mean between — 2 and — f ? 

13. Find the 7th term and the sum of the series 1, 3, 9, • • •. 

14. Find the 6th term and the sum of the series 2, 4, 8, • • • . 

15. What is the geometrical mean between -y/a^h and VoP ? 

16. Find the 7th term and the sum of the series 8, 2, .5, • • •. 

17. Find the 8th term and the sum of the series ^^j, i^, j, • • • . 

18. Find the 7th term and the sum of the series Vi, 2, 2V2, • • •. 

19. Find the 7th term and the sum of the series \/2, V^, Vi, • • • . 

20. Find the 10th term and the sum of the series ■^\-^^ y^^, ^^j, • • •. 

21. Find the 5th term and the sum of the series V2 — 1, 1, 1 + V2, • • •. 

22. The first and sixth terms of a G.P. are 1 and 243. Find the interme- 
diate terms. 

23. Find the 5th term and the sum of the series 
^ :, 6-2V6, 9V3-11V2, ••.. 



V3+ Vii 

24. Insert 3 geometrical means between — and - . 

V3 9 

25. What is the geometrical mean between ~ and ^^ ^? 

X -\- y X -y 

1 - 4 

26. Find the 6th term and the sum of the series , — •\/2, — =, • • .. 

V2 V2 

27. Find the 6th term and the sum of the series -* /- , 1, — --, • • •. 

\3 V2 

28. Find the 5th term and the sum of the series v^, — 1, , • • •. 

29. Find the 6th term and the sum of the series , -» / — , , • • •. 

8 \32 4 



140 QUADRATICS AND BEYOND 

30. The geometrical mean of two numbers is 4 and their sum is 10. Find 
the numbers. 

31. The fourth term of a G.P. is 192, the seventh term is 12,288. Find 
the first term and the ratio. 

32. If the same number be added to or subtracted from each tern^^^^f^ 
G.P., is the resulting series geometrical? 

33. The product of the first and last of four numbers in G.P. is 64. 
Their quotient is also 64. Find the numbers. 

34. The product of four numbers in G.P. is 81. The sum of the second 
and third terms is i. Find the numbers. 

35. If every term of a G.P; be multiplied by the same number m, is the 
resulting series a G.P.? If so, w^hat are the elements? 

36. The sum of three numbers in G.P. is 42. The difference between the 
squares of the first and the second is 60. What are the numbers ? 

37. The difference between two numbers is 48. The arithmetical mean 
exceeds the geometrical mean by 18. Find the numbers. 

38. Four numbers are in G.P. The difference between the first and the 
second is 4, the difference between the third and the fourth is 36. Find the 
numbers. 

39. A ball falling from a height of 60 feet rebounds after each fall one 
third of the last descent. What distance has it passed over when it strikes 
the ground for the eighth time ? 

40. The difference between the first and the last of three terms in G.P. 
is four times the difference between the first and second terms. The sum of 
the numbers is 208. Find the numbers. 

41. An invalid on a certain day was able to take a single step of 18 
inches. If he was each day to walk twice as far as on the preceding day, 
how long before he can take a five-mile walk ? 

42. The difference between the first and the last of four numbers in G.P. 
is thirteen times the difference between the second and third terms. The 
product of the second and third terms is 3. Find the numbers. 

137. I nfinite series. When the number of terms of a G.P. is 
unlimited it is called an infinite geometrical series. 

In the series a, ar, ar^^ • • •, when r > 1, evidently each term is 
larger than the preceding term. The series is then called increas- 
ing. When r < 1, each term is smaller than the preceding term 
and the series is called decreasing. 

T,T . a (1 — r") a ar^ 

Now m any case s = — \ = 

1 — r 1 — r 1 — r 



GEOMETRICAL PROGRESSION 141 

When r > 1, evidently r" becomes very large for large values 
of n. For this case, then, the sum of the first n terms becomes very 
large for large values of n. In fact we can take enough terms 
so that s will exceed any number we may choose. If, however, 
r < 1, as 71 increases in value r" becomes smaller and smaller. In 
fact we can choose n large enough so that r" is as small as we wish, 
or as we say, approaches as a limit. But since r" may be made 
as small as we wish, ar^ also approaches as a limit, and conse- 

quently approaches as a limit. Thus when r < 1 the 

value of the sum of the first n terms approaches as n 

1 — r 

becomes very great. This we express in other words by asserting 
that the sum of the infinite series 

a -\- ar -\- ar^ + ■ "j when r < 1, 

is s^ = 



1-r 

. EXERCISES 

Find the sum of the following infinite series. 

1. 6 + 3 + 1 + .... 
Solution : a =z 6, r = |. 

a 






1-r 



6 


= 5 = 12. 


i-i 


h 


3. 


64 + 8 + 1+.. 


5. 


^ + i^ + :rV + - 


7. 


2 + .5 + .125 + 



2. l + i + i + .... 
4. h + \ + l + -"' 
6. ! + f + -V + --- 

■v/2 
8. V2 + l + -y- + .... 9. (V2 + l) + l + (v^-l) + .... 

10. How large a value of n must one take so that the sum of the first 
n terms of the following series differs from the sum to infinity by not more 
than .001? 

(a) 8 + 4 + 2 + .... 

Solution : a = 8, r = |. 

a ar'* ar** 

8 = = Sao — 



1-r 1-r 1 



8^-8= 

1-r 



142 



QUADRATICS AND BEYOND 



We must find for what value of n the expression 






is less than .001. 



say 



8-2 

2» 



16 

2«* 



'"^ - - 16 1 ^ 

By trial we see that if w = 14 the value of — is , which is less than 

.001 2". 1^2^ 

(b) 27 + 3 + i + . . .. (c) 4 + I + ^ij + • • •. 

(d) 1 + ^1^ + 3,1^ + .... (e) 64 + 16 + 4 + .... 

(f) 100 + 20 + 4 + . . •. (g) 60 + 20 + 6f% . . .. 

11. What is the value of the following recurring decimal fractions? 

(a) .212121.... 

Solution : This decimal may be written in the form 



Here 



(b) .333.... 
(e) .343343 



21 21 21 

100 ' (100)2 ' (100)3"'" • 




21 1- 
100 ' 100 




5 _ « _ .21 _ .21 _ 7 
* 1-r l-.Ol .99 33* 




(c) .717171.... 


(d) .801801.. 


(f) 1.43131.... ^ 


(g) 2.61414.. 



ADVANCED ALGEBRA • 

CHAPTEE XV 
PERMUTATIONS AND COMBINATIONS 

138. Introduction. Before dealing directly with the subject of 
the chapter we must answer the question, In how many distinct 
ways may two successive acts be performed if the first may be 
performed in p ways and the second may be performed in q ways ? 
Suppose for example that I can leave a certain house by any 
one of four doors, and can enter another house by any one of five 
doors, in how many ways can I pass from one house to the other ? 
If I leave the first house by a certain door, I have the choice of 
all five doors by which to enter the second house. Since, how- 
ever, I might have left the first by any one of its four doors, 
there are 4 • 5 = 20 ways in which I may pass from one house 
to the other. This leads to the v,^ 

Theorem. If a certain act may he 'performed in p ways, and 
if after this act is performed a second act may he performed in q 
ways, then the total numher of ways in which the two acts may 
he performed is p • q. 

With each of the p possible ways of performing the first act 
correspond q ways of performing the second act. Thus with all 
the p possible ways of performing the first act must correspond 
q times as many ways of performing the second act. That is, the 
two acts may be performed in ^ • 5' ways. 

It is of course assumed in this theorem that the performance 
of the second act is entirely independent of the way in which the 
first act is performed. 

143 



144 ADVANCED ALGEBRA 

EXERCISES 

1. I have four coats and five hats. How many different combinations 
coat and hat can I wear ? 

Solution : The first act consists in putting on one of my coats, whicli may 
be done in four ways-, the second act consists in putting on one of my 
hats, which may be done in five ways. Thus I have 4 • 5 = 20 different 
combinations of coat and hat. 

2. In how many ways may the two children of a family be assigned to 
five rooms if they each occupy a separate room ? 

3. A gentleman has four coats, six vests, and eight pairs of trousers. In 
how many different ways can he dress ? 

4. I can sail across a lake in any one of four sailboats and row back 
in any one of fifteen rowboats. In how many ways can I make the trip ? 

5. Two men wish to stop at a town where there are six hotels but do not 
wish quarters at the same hotel. In how many ways may they select hotels ? 

6. A man is to sail for England on a steamship line that runs ten boats 
on the route, and return on a line that runs only six. In how many different 
ways can he make the trip ? 

7. In walking from A to B one may follow any one of three roads; in 
going on from B to C one has a choice of five roads. In how many different 
ways can one walk from A to C ? 

139. Permutations. Each different arrangement either of all 
or of a part of a number of things is called a permutation. 

Thus the digits 1, 2 have two possible permutations, taken both 
at a time, namely, 12 and 21. 

The digits 1, 2, 3 have six different permutations when two are 
taken at a time, namely, 12, 13, 21, 23, 31, 32. For if we take 1 
for the first place, we have a choice of 2 and 3 for the second 
place, and we get 12 and 13. If 2 is in the first place, we get 21 
and 23. Similarly, we get 31 and 32. In this process it is noted 
that we can till the first of the two places in any one of three 
ways ; the second place can be filled in each case in only two ways. 
Thus by the Theorem, § 138, we should expect 3-2 = 6 permuta- 
tions of three things taken two at a time. We observe that this 
product 3 • 2 has as its first factor 3, which is the total number of 
things considered. The number of factors is equal to the number 
of digits taken at a time, i.e. two. This leads to the general 



PEKMUTATIONS AND COMBINATIONS 145 

Theorem. Tlie number of permutations of n ohjects taken r 

at a time is , i\ , , -t\ /t\ 

n(n — 1) ■ ■ ' (n — r -\- J^)' v-j 

This is symbolized by P„^ ^. 

This formula is easily remembered if one observes that the first 
factor is 7i, the total number of objects considered, and that the 
number of factors is r, the number of objects taken at a time. 
Thus ■Py3 = 7-6-5. 

We prove this theorem by complete induction. 

Flrstj let r = 1. There are evidently only n different arrange- 
ment of n objects, taking one object at a time, namely (assuming 
our objects to be the first 7i integers), 

1, 2, 3, ..-, 71. 

Let us take two objects at a time, i.e. let r= 2. Since there 

are n objects, we have n ways of filling the first of the two places. 

When that is tilled there are n — 1 objects left, and any one may 

be used to fill the second place. Thus, by the Theorem, § 138, 

there are for r = 2 . . . 

n {n — V) 

different permutations. 

Secondy assume the form (I) for r = m^ 

Pn,m = n(n-l)..-(n-m + l). (1) 

We can fill the first m places in P„ ^ different ways since there 
are that number of permutations of n things taken m at 'a time. 
This constitutes the first act (§ 138). The second act consists in 
filling the m -f 1st place, which may be done in n — m ways by 
using any of the remaining n — in objects. Thus the number of 
permutations of n things taken 7n + 1 at a time is 
Pn,r,^ + i = Pn,nr(n-m)=:=n(n-l){n-2)---(n-m + l)(n-m\ 
which is the form that (1) assumes on replacing m by m + 1. 

Corollary. The number of permutations of n things taken all 

at a time is ^ , -,\ ^ -t » * /o\ 

F^^^=n(n-'l)-'-2-l = n!* (^) 

Taking n = r in (I), we get (2). 

» « / is the symbol for 1 • 2 .3 • 4 •••(»- 1) n, and is read factorial n. 



146 ADVANCED ALGEBRA 

EXERCISES 

1. How many permutations may be formed from 8 letters taken four at a 
time? 

Solution : n = 8, r=»4, n-r + l=6, 

Pg, 4 = 8 • 7 . 6 . 5 = 1680. 

2. In how many different orders may 6 boys stand in a row ? 

3. How many different numbers less than 1000 can be formed from the 
digits 1, 2, 3, 4, 5 without repetition ? 

4. How many arrangements of the letters of the alphabet can be made 
taking three at a time ? 

5. How many numbers between 100 and 10,000 can be formed from the 
digits 1, 2, 3, 4, 5, 6 without repetition ? 

6. How many different permutations can be made of the letters in the 
word compute taking four at a time ? 

7. In a certain class there are 4 boys and 5 girls. In how many orders may 
they sit provided all the boys sit on one bench and all the girls on another ? 

Hint. Use Corollary § 139, and then Theorem, § 138. 

8. I have 6 books with red binding and 3 with brown. In how many ways 
may I arrange them on a shelf so that all the books of one color are together ? 

140. Combinations. Any group of things that is independent of 
the order of the constituents of the group is called a combination. 

The committee of men Jones, Smith, and Jackson is the same 
as the committee Jackson, Jones, and Smith. The sound made by 
striking simultaneously the keys EGrC of a piano is the same as 
the sound made by striking CGE. In general a question involv- 
ing the number of groups of objects that may be formed where 
the character of any group is unaltered by any change of order 
among its constituent parts is a question in combinations. 

Suppose for example that we ask how many committees of three 
men can be selected from six men. If the men are called A, B, C, 
D, E, F, there are, by § 139, 6 -5 • 4 = 120 difPerent arrangements 
or permutations of the six men in groups of three. But the permu- 
tations A, B, C ; A, C, fe ; B, A, C, etc. (3 ! = 6 in all for the men 
A, B, and C), are all distinct, while evidently the six committees 
consisting of A, B, and C are identical. This is true for every 
distinct set of three men that we could select; that is, for the 



PERMUTATIONS AND COMBINATIONS 147 

six different permutations of any three men there is only one 
distinct committee. Hence the number of committees is one sixth 

the total number of permutations, or -^' 
This leads to the general 

Theorem. The number of comhinations of n things taken r at 

a time is . -,^ , . ^x 

n(n — I)--- (n — T-\- 1) 

This is symbolized by C„^ ^. 

The number of permutations of n things taken r at a time is 

p^^^ = n(n - 1) • ' • (r, — r + 1). 

In every group of r things which form a single combination 
there are (Cor., p. 145) r ! permutations. Thus there are r ! times 
as many permutations as combinations. That is, 

r -?Ji^- n{n-l)-'-{n-r-\-l) 
"''•""/•!" r\ ' ^^ 

This formula is easily remembered if one observes that there 
is the same number of factors in the numerator as in the denomi- 
nator. Thus 

10- 9- 8 
^10.3- -^.2-3 
Corollary. C ^= C 



n, n — r 



Multiplying numerator and denominator of (I) by (n — r)!, 
_ n(n-l)-'-(n-r + l)(n-r)-'-2-l 
"•'•" rl(n-ry. 



_ n{n-l)---{r-\-l) 
(n — r)\ 

— ^ (^ ~ 1) • • • [^ ~ (^ ~ ^) + ^] 

(n — r)\ 



n.n — r' 



This corollary saves computation in some cases. For instance, if we wish to 
compute Ci9, ir, it is more convenient to write Ci9, 17= Ci9, 2 = ^ , = 171 than the 
expression for C'lo, 17. 



148 ADVANCED ALGEBRA 

EXERCISES 

1. How many committees of 5 men can be selected from a body of 10 
men three of whom can serve as chaii'man but can serve in no other capacity ? 

Solution : There are 7 men who may fill 4 places on the committee. 



^' 1.2.3-4 



There are 3 men to select from for the remaining place of chairman, 
and the selection may be made in 3 ways. Thus the committee can be 
made up in 3 • 35 = 105 ways. 

2. How many distinct crews of 8 men may be selected from a squad of 
14 men ? 

3. How many distinct triangles can be drawn having their vertices in 
10 given points no three of which are in a straight line ? 

4. How many distinct sounds may be produced on 9 keys of a piano by 
striking 4 at a time ? 

5. In how many ways can a crew of 8 men and a hockey team of 5 men 
be made up from 20 men ? 

6. In how many ways may the product a-b • c • d- e -f be broken up 
into factors each of which contains two letters? 

7. If 8 points lie in a plane but no three in a straight line, how many 
straight lines can be drawn joining them in pairs? 

8. How many straight lines can be drawn through n points taken in 
pairs no three of which are in the same straight line ? 

9. Seven boys are walking and approach a fork in the road. They 
agree that 4 shall turn to the right and the remainder turn to the left. In 
how many ways could they break up ? 

Solution : The number of groups of 4 boys that can be formed from the 

CV 4 = = o5. 

4! 

For each group of 4 boys there remains only a single group of 3 boys. 
Thus the total number of ways in which tliQ party can divide up is 
precisely 35. 

10. If there are 12 points in space but no four in the same plane, how 
many distinct planes can be determined by the points? 

Hint. Three points determine a plane. 

11. Eight gentlemen meet at a party and each wishes to shake hands 
Tfith all the rest. How many hand shakes are exchanged ? 



I 



PERMUTATIONS AND COMBINATIONS 149 

12. In how many ways can a baseball team of 9 men be selected from 
14 men only two of whom can pitch but can play in no other position ? 

13. How many baseball teams can be selected from 15 men only four of 
whom can pitch or catch, provided these four can play in either of the two 
positions but cannot play elsewhere ? 

14. Two dormitories, one having 3 doors, the other having 6 doors, stand 
facing each other. A path runs from each door of one to every door of the 
other. How many paths are there ? 

15. Show that the number of ways in which p -{- q things may be divided 

into groups of p and q things respectively is ^ ^^• 

p\q\ 

16. Out of 8 consonants and 3 vowels how many words can be formed 
each containing 3 consonants and 2 vowels ? 

17. A boat's crew consists of 8 men, three of whom can row only on one 
side and two only on the other. In how many ways can the crew be arranged ? 

18. A pack of cards contains 62 distinct cards. In how many different 
ways can it be divided into 4 hands of 13 cards each ? 

19. Five points lie in a plane, but no three in any other plane. How 
many tetrahedrons can be formed with these points taken with two points 
not in the plane ? 

141. Circular permutations. By circular permutations we 
mean the various arrangements of a group of things around a 
circle. 

Theorem. The number of orders in which n things may he 
arranged in a circle is (n—l) !. 

Suppose A is at the point at which we begin to arrange the 
digits 1, 2, 3, • • •, n. Suppose we start our arrangement of digits 
at A with a given digit a. We have then 
virtually n—l places to fill by the remaining 
n—l digits. Thus we get {n—l)\ (p. 145) 
permutations of the n digits keeping a fixed. 
But suppose we start our arrangement, that is, 
fill the place at A with any other digit, as h, 
and the remaining places in any order what- 
ever. If we now go around the circle till we 
come to the digit a, the succession of digits from that point 
around the circle to a again must be one of the {n —1)\ orders 




150 ADVANCED ALGEBRA 

which we obtained when we took a as the initial figure. Thus 
the only distinct orders in which the n digits can be arranged 
on a circle are the (n —1)\ permutations we obtained by filling 
the first place with a. 



EXERCISES 

1. In how many orders can 6 men sit around a circular table ? 
Solution : 






n = Q, n - 1 = 5, (n - 1)! = 5 ! = 120. 

2. In how many ways can 8 men sit around a circular table ? 

3. In how many ways may the letters of live be arranged on a circle ? 

4. In how many ways may the letters of permutation be arranged on a 
circle ? 

5. In how many ways can 4 men and 4 ladies sit around a table so that 
a lady is always between two men ? 

6. In how many ways may 4 men and their wives be seated around a table 
so that no man sits next his wife but the men and the women sit alternately ? 

7. In how many ways can six men and their wives be seated around a 
table so that each man sits between his wife and another lady ? 

8. In how many ways can 10 red flowers and 5 white ones be planted 
around a circular plot so that two and only two red ones are adjacent ? 

142. Theorem. The number of permutations of n things of 

. n! 
which p are alike, taken all together y is -^• 

If all the things were different, we should have n ! permutations. 
But since p of the n things are alike, any rearrangement of those 
p like things will not change the permutation. Eor any fixed 
arrangement of the n things there are p ! different arrangements 

of the p like things. Thus — : of the n\ permutations are iden- 

tical, and there are only — '- distinct permutations of the n things 
^p of which are alike. ' 



Corollary. If of n things p are of one kind, q of another 

n f 
kind, r of another, etc., then there are — ; — -^— — permutations 

of the n things taken all at a time. -^ ' ^' ' ^ 



PERMUTATIONS AND COMBINATIONS 151 

EXERCISES 

1. How many distinct arrangements of the letters of the word Cincinnati 
are possible ? 

Solution : There are in all 10 letters, of which 3 are i, 2 are c, and 3 are n. 
Thus the number of arrangements is 

10! _ X.2.^-^-5-^-7-8-9.10 
3!3!2!" ;.^.3X-^-^;-;Z 
= 2.5-7.8.9.10 = 50,400. 

2. How many distinct arrangements of the letters of the word parallel 
can be formed ? 

3. How many signals can be made by hanging 15 flags on a staff if 2 
flags are white, 3 black, 5 blue, and the rest red ? 

4. How many signals can be made by the flags in exercise 3 if a white 
one is at each extreme ? 

5. How many signals can be made by the flags in exercise 3 if a red flag 
is always at the top ? 

6. Would 3 dots, 2 dashes, and 1 pause be enough telegraphic symbols 
for the letters of the English alphabet, the numerals, and six punctuation 
marks ? 



CHAPTEE XVI 
COMPLEX NUMBERS 

143. The imaginary unit. When we approached the solution 
of quadratic equations (p. 52) we saw that the equation x^ = 2 
was not solvable if we were at liberty to use only rational num- 
bers, but that we must introduce an entirely new kind of number, 
defined as a sequence of rational numbers, if we wished to solve 
this equation. The excuse for introducing such numbers was not 
that we needed them as a means for more accurate measurement, 
— the rational numbers are entirely adequate for all mechanical 
purposes^ — but that they are a mathematical necessity if we 
propose to solve equations of the type given. 

A similar situation demands the introduction of still other 
numbers. If we seek the solution of 



£C2=-1, 



(1) 



we observe that there is no rational number whose square is — 1. 
Neither can we define V— 1 as a sequence of rational numbers 
which approach it as a limit. We may write the symbol V— 1, but 
its meaning must be somewhat remote from that of V2, for in 
the latter case we have a process by which we can extract the 
square root and get a number whose square is as nearly equal to 
2 as we desire. This is not possible in the case of V^l. In fact 
this symbol differs from 1 or any real number not merely in 
degree but in kind. One cannot say V— 1 is greater or less than 
a real number, any more than one can compare the magnitude of 
a quart and an inch. 

V — 1 is symbolized by I and is called the imaginary unit. The 
term "imaginary" is perhaps too firmly established in mathe- 
matical literature to warrant its discontinuance. It should be 
kept in mind, however, that it is really no more and no less 

152 



COMPLEX NUMBERS 153 

imaginary than the negative numbers or the irrational numbers 
are. So far as we have yet gone it is merely that which satis- 
fies equation (1). When, however, we have defined the various 
operations on it and ascribed to it the various characteristic 
properties of numbers we shall be justified in calling it a 
number. 

Just as we built up from the unit 1 a system of real numbers, 
so we build up from V— 1 = i a system of imaginary numbers. 
The fact that we cannot measure V— 1 on a rule should cause 
no more confusion than our inability exactly to measure -y/2 on a 
rule. Just as we were able to deal with irrational numbers as 
readily as with integers when we had defined what we meant 
by the four operations on them, so will the imaginaries become 
indeed numbers with which we can work when we have defined 
the corresponding operations on them. 

144. Addition and subtraction of imaginary numbers. We 
write 

= 0i, 

i -\- i = 2 ij 
i -{- i-\ -{- i = ni. (I) 



Also just as we pass from a rational to an irrational multiple 
of unity by sequences, so we pass from a rational to an irrational 
multiple of the imaginary unit. Thus we write a V— 1, or ai, 
where a represents any real number. Consistently with § 76 we 
write 

^ V-a2 = ^ Va2.(_l) = ± V^ . V^ = ±a V^ = ± ai. (II) 

We speak of a positive or a negative imaginary according as 
the radical sign is preceded by a positive or a negative sign. 
We also define addition and subtraction of imaginaries as 

follows : 

ai ± bi = (a ±b) i, (III) 



where a and b are any real numbers. 



154 ADVANCED ALGEBRA 

Assumption. The commutative and associative laws of multi- 
plication and addition of real numbers, § 10, we assume to hold 
for imaginary numbers. 

145. Multiplication and division of imaginaries. We have 
already virtually defined the multiplication of imaginaries by 
real numbers by formula (I). Consistently with § 76 we define 

V^ . V^ = ii = i^ = - 1. 

Thus V— a • V— b = Va • V^ i i = -\/ab • (— 1) = — -Vab. 

The law of signs in multiplication may be expressed verbally 
as follows : 

The product of imar/inaries with like signs before the radical 
is a negative real number. The product of imaginaries with 
unlike signs is a positive real number. 

For instance, - yP^ - V^^ = - 2 • 3 • i2 = 6. 

We also note that 

i2 = — 1, i3 = — i, i* = 1, i^ =r i, . . . , 
And, in general, Hn + k _ ik^ /c = 0, 1, 2, 3. 

We define division of imaginaries as follows : 

/ I r Vo^ • i la 

■Vb-i ^b 

In o perating with imaginary numbers, a number of the form 
V— a should always be written in the form Va i before per- 
forming the operation. This avoids temptation to the following 

error: 

V- a • V- b = V(- a) ■(-b) = -slab. 

EXERCISES 

Simplify the following : 

1. V^8 ■ yT^. 

Solution : V^^ • V^^ = VS • i . V2 • i = V2 • 8 . {2 = 4 . (_ 1) = _ 4. 

2.1. 

1 2^ /^ ■^— % 

Solution : — = - = = = — i. 

i^ i8 (i4)2 \ 



COMPLEX NUMBEKS 



165 



3. 


i". 


6. 


V-36. 


9. 


V- X2«. 


12. 


V2 V- 8. 


15. 


1 


18. 


V-6 



4. 124. 


5. ii3. 


7. V-64. 


8. 2i'Si. 


10. V-Sx^a^. 


11. V-x2. 


13. V-2V-6. 


14. V-3% 


16.1. 


17.. f^. 

V-2 


19. V-i2. 


20. V-i^. 



146. Complex numbers. The solution of the quadratic equation 
with negative discriminant (p. 71) affords us an expression which 
consists of a real number connected with an imaginary number 
by a H- or — sign. Such an expression is called a complex number. 
It consists of two parts which are of different kinds, the real 
part and the imaginary part. Thus 6 + 4 z means G I's + 4 ^'s. 
Obviously, to any pair of real numbers (x, y) corresponds a complex 
number x + ty, and conversely. 

147. Graphical representation of complex numbers. We have 
represented all real numbers on a single straight line. When we 
wished to represent two numbers simultaneously, we made use of 
the plane, and assumed a one-to-one correspondence between the 
points on the plane and the pairs of numbers (cc, y). The general 
complex number x -f- iy depends 
on the values of the independent 
real numbers x and y, and may 
then properly be represented by 
a point on a plane. We repre- 
sent real numbers on the X axis, 
imaginary numbers on the Y axis, 
and the complex number x + iy 
by the point (x, y) on the plane. 
Thus the complex numbers 6 + ^ 3, 
— 4 -f 1 4, 7 — i 5, — 2 — 1 4 are represented by points on the plane 
as indicated in the figure. 

148. Equality of complex numbers. We define the two com- 
plex numbers a -f ib and c -{- id to be equal when and only when 
a = c and h — d. 











y> 


k 


















- 


4+ 


i4 














































6f 


i?, 



















































































































X 
































































-2- 


i4 






































































7- 


15 




























rn 





156 ADVANCED ALGEBRA 

Symbolically a + ii ^ c + id 

when and only when a = c, b = d. 

The definition seems reasonable, since 1 and i are different in 
kind, and we should not expect any real multiple of one to cancel 
any real multiple of the other. 

Similarly, if we took not abstract expressions as 1 and i for 
units but concrete objects as trees and streets, we should say that 

a trees + b streets = c trees + d streets 
when and only when a = c and b = d. 

Principle. When two numerical expressions involving imagi- 
naries are equal to each other, we may equate real parts and 
imaginary parts separately. 

The graphical interpretation of the definition of equality is that equal complex 
numbers are always represented by the same point on the plane. 

From the definition given we see that a -\-ib = when and 
only when a = b = 0. 

Assumption. We assume that complex numbers obey the com- 
mutative and associative laws and the distributive law given in 
§ 10. We also assume the sa.me rules for parentheses as given 
in § 15. 

This assumption enables us to define the fundamental opera- 
tions on complex numbers. 

149. Addition and subtraction. By applying the assumptions 
just made we obtain the following symbolical expression for the 
operations of addition and subtraction of any two complex num- 
bers a -{-ib and c -{- id: 

a -{• ib ± (c -\- id) = a ± c -{- i(b ± d). 

Rule. To add (subtract) complex numbers, add (subtract) the 
real and imaginary parts separately. 

150. Graphical representation of addition. We now proceed 
to give the graphical interpretation of the operations of addition 
and subtraction. 



COMPLEX NUMBERS 



167 



Theorem. The sum of two numbers A = a-\- ib and B = c -\- id 
is represented hy the fourth vertex of the parallelogram formed 
on OA and OB as sides. 

Let OASB'hQ ?i parallelogram. Draw 
ES _L OE, AH A. ES, BD (= d) _L OE. 
A AHS = A ODE since their sides are 
parallel, and OB — AS. 

Thus 



Thus 



DB = 


: HS = d, 


0D = 


AH==c. 


ES = 


EH + HS 



Y 


1 


d A 


7 


S 
H 






b 







D 

—a, > 


JF E X 



b + d, 



OE = OF + FE = a -\- c, 

and S has coordinates (a -\- c, b -{- d) and represents the sum of 
A and B, by § 149. 

EXERCISES 

1. The difference A — B of two numbers A = a + ib and B = c + id is 
represented by the extremity D of the line OD drawn from the origin par- 
allel to the diagonal BA of the parallelogram formed on OB and OA as 



2. Represent graphically the following expressions. 

(a) 1 + i (b) - 4 - 2 i ■ 

(c) 6 - i. (d) - 8 + 4 i. 

(e) 2 + 4 I (f) (1 + i) + (2 + i)' 

(g) {2-i)-{6-Si). (h) (l_i)-(l_2i). 

(i) (2 + 4i)-(l-3i). (j) 4(l + i)-2(2-3i). 

(k) (6-2i) + (2 + 3i). (1) (5 + 3i) + (-l-6i). 

151. Multiplication of complex numbers. The assumption of 
§ 148 enables us to multiply complex numbers by the following 

Rule. To multiply the complex number a -\- ib by c ■\- id, pro- 
ceed as if they were real binomials, keeping in mind the laws for 
multiplying imaginaries. 

Thus a -\-ib 

c -f- id 

ac + icb -f- iad -\- (i)* bd = ac — bd + i(cb -f- ad). 



158 ADVANCED ALGEBRA 

152. Conjugate complex numbers. Complex numbers that differ 
only in the sign of their imaginary parts are called conjugate com- 
plex numbers, or conjugate imaginaries. 

Theorem. The sum and the product of conjugate complex 
numbers are real numhers. 

Thus a -\- lb -\- a — ib = 2 a, 

(a + ib) (a - ib) =a^ + b\ 

153. Division of complex numbers. The quotient of two com- 
plex numbers may now be expressed as a single complex number. 

EuLE. To express the quotient — in the form x + ^y, 

rationalize the denominator^ using as a rationalizing factor 
the conjugate of the denominator. ^_^ 

™,, a -{- ib a -\- ib c — id 

Thus — = ~ ^^—^ 

c -\- id c -\- id c — id 

_ ac -\- bd — i {ad — be) 

~ c' + d^ 

ac -{- bd .ad — be 

^ > + d' ~ "" VT^' ^^ 

We have now defined the fundamental operations on complex 
numbers and shall make frequent use of them. If the question 
remains in one's mind, "After all, what are they? " the answer is 
this : " They are quantities for which we have defined the funda- 
mental operations of numbers and, since they have the properties 
of numbers, must be called numbers, just as a flower that has all 
the characteristic properties of a known species is thereby deter- 
mined to belong to that species." Furthermore, our operations 
have been so defined that if the imaginary parts of the complex 
numbers vanish and the numbers become real, the expression 
defining any operation on complex numbers reduces to one defin- 
ing the same operation on the real part of the number. Thus in 
(1) above, if b = d = 0^ the expression reduces to 

a _a 

a c 



COMPLEX NUMBERS 159 



EXERCISES 

Carry out tlie indicated operations. 

1. (2 + V"r2)(4+V^^). 
Solution : 2 + V^^ = 2 + ■v/2(-l) = 2 + z V2 
4 + V:^ =:4+V5(-l) = 4 + t V6 



8 - VlO + i4V2 + i2\/5 

2. 5 - V2 - i Vs. 

Solution : 

5^ ^ 5(V2 + zV3) ^ 5V2 + Z5V3 _ /g + i Vs 

\/2-iV3 (V2-iV3)(V2 + iV3)~ 2 + 3 

3. (l + i)l 4. (l + i)3. 
Hint. Develop by the binomial theorem. 

5. {a + ib)K 6. (V^ + v^ir^)^ 

V7. (x + %)2 8. (x + i2/)2 + (X - %)2. 

9. vT+l.Vnri. 10. (V3 + iv^)(V2 + iV3). 

11. (VTT~i + Vr^y. 12. (aV6 + icVd)(aV6-icVd). 

13. (Va + i V6) ( Va - i Vb). 14. (2 V7 + i3 Vs) (3 V7 - ilOV2), 

15. i±i^. 16. l±i. 17. ^ 



l-iV3 1-i V2+V-1 

18. ^ . 19. 11^. 20. (^l±i^)'. 

l + V-3 (l + i)3 \ 2 / 



2„ g + i Vl - a2 
a — i Vl — a^ 

24. ;^-^- 25. -^1— _. 26. ^ + ^p^ . 

27. ^^ + ^^'^ 28 (l±i^y. 29. ^ + ^^ _j_ c + id 



V3-iV2 \2/ a-ib c -id 

30. ?? 31. — ?i=. 32. -V+ 1 



4 + 7V^6 i + 3v^^. (1 + 0' a-*y 

33 VI + « + ^ Vl- <^ _ Vl - g + i Vl + g 
Vi -j- g — i VT — g Vl — a — i Vl + a 



160 ADVANCED ALGEBRA 

34. Find three roots of the equation x^ — 1 = and represent the roots 
as points on the plane. 

35. Find four roots of the equation x* — 1 = and represent the roots as 
points on the plane. 

36. Find six roots of x^ — 1 = and represent the roots as points on the 
plane. Show graphically that the sum of the six roots is zero. 

37. Find three roots of x^ — 8 = and represent the roots as points on 
the plane. Show graphically that the sum of the three roots is zero. 

154. Polar representation. The graphical representation of 
complex numbers given in § 147 gives a simple graphical inter- 
pretation of the operations of addition and subtraction, but the 
graphical meaning of the operations of multiplication and divi- 
sion may be given more clearly in another manner. We have 
seen that we may represent x -f iy by the point P (x, y) on the 
plane, Represent the angle between OP and the X axis by/^. 
This angle is called the argument of the complex number x -f iy. 

Eepresent the line OP by p. This is called 
^ , . , the modulus of ic -f iy. Then from the figure 

a; = p COS B, (1) 

— ^ y = psmO, (2) 

x' + f = p\ (3) 

Hence the complex number x + iy may be written in the form 

X -\- iy = p (cos -}- i sin 0)j (4) 

■when the relations between x, y and p, are given by (1), (2), and 
(3). A number expressed in this way is in polar form, and may 
be designated by (p, 6). We observe that a complex number 
lies on a circle whose center is the origin and whose radius 
is the modulus of the number. The argument is the angle 
between the axis of real numbers and the line representing the 
modulus. 

155. Multiplication in polar form. If we have two numbers 
p (cos 6 -\-i sin^) and /o'(cos 6' -\- i sin ^'), we may multiply them 
and obtain 



COMPLEX NUMBERS 161 

p (cos 6 + isin 0)p'(Gos 0' + i sin 6') 

= pp' [(cos cos 0' — sin 6 sin d') 
+ i (sin $ cos d' + cos sin ^')] 
By the addition theorem ,r //i , /if\ , • • /h , /if\n /-i\ 

in Trigonometry = PP C^^^ (^ + ^') + ^ sm (d + d')] (1) 

= R (cos © + * sin ©) . (2) 

In this product pp' is the new modulus and 0-^0' the new 
argument. We may now make the following statement: The 
product of the two numbers p (cos 6 -\- i sin 6) and p'(cos 6' + * sin 0') 
has as its modulus pp' and as its argument + 0'. Thus the 
product of two numbers is represented on a circle whose radius 
is the product of the radii of the circles on which the factors are 
represented. The argument of the product is the sum of the 
arguments of the factors. 

156. Powers of numbers in polar form. When the two factors 

of the preceding section (p, 0) and (p', 6') are equal, that is, 

when p = p' and 6 = 6', the expression (1) assumes the form 

[p (cos e + i sin 0)^ = p^ (cos 2 ^ + t sin 2 0). (1) 

This suggests as a form for the nth. power of a complex number 
[p (cos d -\-i sin $)']" = p« (cos nO -\- i sin nO). (2) 

The student should establish this expression by the method of 
complete induction. The theorem expressed by (2) is known as 
DeMoivre's theorem. Stated verbally it is as follows : The modulus 
of the nth. power of a number is the nth power of its modulus. The 
argument of the nth power of a number is n times its argument. 

EXERCISES 

Plot, find the arguments and moduli of the following numbers and of 



eir products. 
1. 1 + iVS, V3 + i. 
Solution : 

Let ^/S -\- i=p (cos ^ + i sin 5), 
1 4- i V3 = /(cos d' + i sin 6^. 


2. 

30°; 


Y. 


B 

Ay ! i 1 ! 


Then by (1), (2), (3), § 154, p = 2 ; p' = 
1 = 2 sin d, hence 6 = 





'--- -1— J 1 X 


1 = 2 cos d", hence 6' = 


60°. 




= 30°, ^' = 60° 



162 ADVANCED ALGEBRA 

Thus if the product has the form R{cos@ + isin©), we have by § 1( 
R = ppr = 4, © = ^ + ^' = 90°. 



2. l+i,2 + i. 


3. (l-i)3. 


4. 3 + 3t, 2^iVl2. 


5. 2i, l-iV3. 


'4*'-^)- 


7.-1 + 1,-2-2. 


o , . V2 iV2 
^- '^'2 2 • 


'-2+ 2 ' 2 + 2 



10. [2 (cos 15° + i sin 15°)]8. 11. [i (cos 30° + i sin 30°)]4. 

12. [| (cos 120° + i sin 120°)]2. 13. [2 (cos 135° + i sin 135°)]*. 

14. [f (cos 180° + isin 180°)] 3. 15. [f (cos 315°+ isin 315°)]2. 

157. Division in polar form. If we liave, as before, two com- 
plex numbers in polar form (p, 6) and (p', $'), we may obtain their 
quotient as follows. 

p (cos -{- i sin $) 
p' (cos 0' -^ isin 0') 

_ pp' (cos -{- i sin 0) (cos 0' — i sin 0') 
~ p'2 (cos ^' + 2 sin $') (cos ^' — i sin ^') 
^ pp' [cos (6 - ^0 + i sin (^ - ^')] 
~ p'2(cos2^' -f sin^^') 



Rationalizing, 



§ 152 and § 153, 



Since sin2 + cos2 6=1, = -^ [cos (0 — $') + i sin (^ — 0')'] 

= i2 (cos^Vf i sin ©). 

We may now make the following statement: The, quotient of 
two complex numbers has as its modulus the quotient of the moduli 
of the factors, and as its argument the difference of the arguments 
of t/ie factors. 

158. Roots of complex numbers. We have seen that the square 
of a number has as jts modulus the square of the original modulus, 
while' the argument is twice the original argument. 

Thjs would suggest th^-t the square root of any number, as (p, 0), 

' : ' - 9 

would have Vp as its modulus and - as its argument. Since 

'^ ^ 

every yeal number has two square roots, we should expect the 

same fact to hold liere. Consider the two numbers 



COMPLEX NUMBERS 



163 



Vp /^cos I + i sin I j and Vp cos (^ -f 180° j + i sin ( | + 180° j j , 

where Vp is the principal square root of p(§ 72). The square of 
the first is (p, ^), by § 155. That the square of the second is the 
same is evident if we keep in mind the fact that 

cos (0 + 360°) = cos e 
and sin (^ + 360°) = sin a 

Thus V/> (cos e + i sin 6) 



Vp ( cos - + i SlUv- 



or 



Vp 



+ 180° -f j sin/ ^ + 180 



■)]■ 



The graphs of these two numbers are situated at points sym- 
metrical to each other with respect to the origin. 

We may obtain as the corresponding expression for the higher 
roots of complex numbers the following : 



■v^/>(cos^ + ?:sin^)= V^ I ( 



(9 + /c360°\ . . /^ + A;360' 

cos I + I sin 

j_ \ 71 / \ n 



where for a given value of /i, k takes on the values 0, 1, • • •, t^ — 1, 
and where VP indicates the real positive 7ith root of p. 



EXERCISES 

Perform the indicated operations and plot 
1. 2-2V3i--l+i. 
Solution : 

Let \-\-i — p (cos + i sin ^), 

2 -- 2 V3 i = /(cos e'-\- i sin 6"). 
Then p = \/12 + 1- = V2, 

/ = V22 + (-2V3)^ = 4. 



By (1) and (2), § 154, 

sin ^ = cos ^ = — ^ , hence 6 = 46°. 
V2 




'4; 



'^-^ 



(a-iaW; 



164 

Similarly, 



ADVANCED ALGEBliA 



300° 



sin^ = 


- 


2V3 
4 


V3 
2 


cos^ = 


f 


= 1, hence d' - 



2 _ 2 V3 i 4 (cos 300° + i sin 300°) 

Thus = i2(cos© + isin©)=— 7^- -— . . ,^^ ■ 

1+i ^ ' V2(cos46°4- tsin45°) 

Hence by § 157, iJ = -^ = 2v^, © = 300° - 45° = 255» 

V2 



/ 



2. V-2 + 2V3i 
Let -2 + 2 V3i = p(cos^4'isin^). 
Then (§ 154) /> = 4, cos^= - f = -^, 



-2+i2A^ 



and 



120^= 



V-2 + 2V3i = V4 (cos 120° + i sin 120°) 

T5 «i^Q /if /120° + A:360°\ 
By §158, =V4|cos( ^ j 



l+iVs" 






. . /120° + A;360°\'] 
.sm( ^)J 



+ ^ 
(where A; = or 1) 

= 2 (cos 60° + i sin 60°) = 1 + i V3, when k-0. 
= 2 (cos 240° + i sin 240°) = - 1 - i V3, when A; = I 

. 3. VV2 + i V2. 



4. V 




1 + i -^ 1 - i. 
1 V3 



6. 



1 iV3 

2 2 



^1 + i. 



8 -l-i-^-^l + l N 

2 2 ■ 4 4* 

10. -2\/2-2 V2i-T--2 + 2V3i 



7. i + i^:t_j:Jl 
2 2 

9. 2-iVl2-4-3 + 3i. 

11. \^l(cosl5°+isinl5°). 
Solution : 

"/TT ^:,o . - - -. ^Q^ '/tF /15° + fe • 360°\ . . . /16°+ fc • 360°\ 1 

Vl(cosl5°+ ism 15°) = VI cosi ■ ) + ism I 1 

(where fc = 0, 1, or 2). 

1 (cos 5° + t sin 5°), when A; = 0, 
1 (cos 125° + i sin 125°), when A; = 1, 
1 (cos 245° + i sin 245°), when A; = 2. 

12. Yi. 

13. ^^161. 

14. •V'2-f- 2\/3i. 

15. v'cos330°+ tsin330<». 




COMPLEX KUMBERS 



165 



16. V27(cos75°+ism75°). 17. Vl6 (cos 200° + i sin 200°). 

18. Solve the following equations and plot their roots. 

(a) x5 - 1 = 0. 

Solution : x^ = 1, or a; = VT. 

Let 1 = 1 + • i = p (cos ^ + i sin 6). Then p = l, 6 = 0°. 



X = V 1 (cos 0° 4- i sin 



0°) = Vircos/* 



0° + A; . 360' 



(where k takes on the values 0, 1, 2, 3, 4) 

f cos 0° + i sin 0° = 1, when k = 0, 
cos 72° + i sin 72°, when A; = 1, 
cos 144° + i sin 144°, when fe = 2, • 
cos 216° + t sin 216°, when A; = 8, 
cos 288° + i sin 288°, when A; = 4. 

These numbers we observe lie on a circle of 
unit radius at the vertices of a regular pentagon. 



(b) a;4 - 1 = 0. 
(e) x6 - 1 = 0. 

J 



(C) X3-1 = 0. 
(f ) X8 - 1 = 0. 




C?x~^ <Jtf^.C<. 



(d) x5 - 32 = 0. 
(g) x8 -- 27 = 0. 



J^l^^ 



y. 



O^ 



CHAPTEE XVII 
THEORY OF EQUATIONS 

159. Equation of the nth degree. Any equation in one variable 
in whicli the coefficients are rational numbers can be put in the 

form 

f (x) =aoX'^ + a^x--' -}--'• + a, = 0, (1) 

where Gq is positive and a^, • • •, a„ are all integers. 

The symbol /(x) is read "foix" and is merely an abbreviation for the right- 
hand member of the equation. Often we wish to replace x in the equation by 
some constant, as a, — 2, or 0. We may symbolize the result of this substitution 
by/(a),/(-2),or/(0). 

Thus f{b) = a(fi» + aift"" ^ + ^ • • + a„ . 

We symbolize other expressions similarly by <f) (x), Q{x), etc. 

When we speak of an equation we assume that it is in the form 
of (1). This equation is also written in the form 

a;" + M"-i + ---^„ = 0, (2) 

where bi = —> b^ = —> •• • K — ~' 

The 5's are integers only when ai, «2j • • -, «„ are multiples of Qq. 

160. Remainder theorem. We now prove the following impor- 
tant fact. 

Theorem. When f(x) is divided hy x — c, the remainder is 
f(x) with c substituted in place of the variable. 

Divide the equation (1) by x — c. Let R be the remainder, 
which must (§ 26) be of lower degree in x than the divisor ; 
that is, in this case, since x — c is the divisor, R must be a con- 
stant and not involve x at all. Let the quotient, which is of 
degree n ~ 1 in x, he represented by Q (x). 

160 



THEORY OF EQUATIONS 167 

Then i^ = Q(x)+ -5— 

X — C ^ ^ X — G 

Clearing of fractions, 

f(x)= Q(x){x-g)+R. 

But since this equation is an identity it is ahvays satisfied 
whatever numerical value x may have (§ 53). 

Let X = c. 

Then /(c) = a,c- + %c"-i + ... + «„= Q (c) (c - c) + i2. 

But since c — c = 0, Q(c) (c — c) = 0, and 

R = aoc" + ^ic"- 1 H h a„ =/(c)- 

Corollary. If c is a root of f(x) = 0, then x — c is a factor 
of the left-hand member. , 

For if c is a root of the left-hand member, it satisfies that 
member and reduces it to zero when substituted for x. Thus by 
the previous theorem we have, since 

aoC* H- o^ic"-^ H a^ = R = 0, 

f{x)=Q(x)(x-c). 

161. Synthetic division. In order to plot by the method of 
§ 103 the equation 

y = a^x"" + a^x""- 1 H \- a^, 

when the a's are replaced by integers, we should be obliged 
laboriously to substitute for x successive integers and find corre- 
sponding values of y, which for large values of n involves con- 
siderable computation. We can make use of the preceding theorem 
to lighten this labor. The object is to find, with the least possible 
computation, the remainder when the polynomial f(x) is divided 
by a factor of form x — c, which by the preceding theorem is the 
value of f(x) when x is replaced by c, that is, the value of y 
corresponding to x = c. For illustration, let 

f(x) = 2 x^ - 3 a;« -h ?c2 - a: - 9 and c = 2. 



168 ADVAIJCED ALGEBKA 

By long division we have 

a;-2| 2x^-3a;^H- ^' - x- 9 |2a;« + o^' + 3a; + 5 

1 x» + a;2 
lx3-2a;2 



3a;2- X 

3a;2-6x 



5a;- 9 
5a; -10 

+ 1 

We can abbreviate this process by observing the following 
facts. Since x is here only the carrier of the coefficient, we may 
omit writing it. Also we need not rewrite the first number of 
the partial product, as it is only a repetition of the number 
directly above it in full-faced type. Our process now assumes 
the form 

1- 21 2-3 + 1-1- 9 |2 + l + 3 + 5 

:l-4 
+ 1 -? / 

/ -2 



+ 3 ,' 
-6 



+ 5 



10 



+ 1 

Since the minus sign of the 2 changes every sign in the partial 
product, if we replace — 2 by + 2 we may add the partial prod- 
uct to the number in the dividend instead of subtracting. This 
is also desirable since the number which we are substituting for 
X is 2, not — 2. Thus, bringing all our figures on one line and 
placing the number substituted for x at the right hand, we have 

2-3 + 1-1- 9[2 

+ 4 + 2 + 6 + 10 
2+1+3+5+ 1 



THEORY OF EQUATIONS 169 

We observe that the figures in the lower line, 2, 1, 3, 5, up to 
the remainder are the coefficients of the quotient 2 x^ -{- x^ -\- S x -\- 5. 

EuLE FOR SYNTHETIC DIVISION. Write the coefficients of the 
polynomial in order, supplying when a coefficient is lacking. 

Multiply the number to he substituted for x by the first coeffi- 
cient, and add (algebraically) the product to the next coefficient 

Multiply this sum by the number to be substituted for x, add to 
the next coefficient, and proceed until all the coefficients are used. 
The last sum obtained is the remainder and also the value of 
the polynomial when the number is substituted for the variable. 

162. Proof of the rule for synthetic division. This rule we now 
prove in general by complete induction. Let the polynomial be 

a^x"" 4- a^x^-^ + a^x""-^ -\ h «„. 

Let the number to be substituted for x be a. 

First. Let n = 2. Carry out the rule on a^x"^ + a^x + a^. 

We have , , . 

+ aQ(X + {apa + a\) cc 

<^oi + «o«^ + «i, + {a^a -|- a.i) a + ^2 = a^a^ + a^a, + a^. 

Second. Assume the validity of the rule for n = m, and prove 

that its validity f or ti = m + 1 follows. Assume then that the rule 

carried out on 

f{x) = a^x^ + a^x^-^ + ••• + «„ 

affords the remainder 

a^a"^ + ai^"*-! ^ -\- a^ =f((x)' 

Now the polynomial of order w + 1 is 

^0^"'+^ 4- aix"" H \-a^x-i- a^ + j = x •f(x) + a^^^ 

Hence the next to the last remainder obtained by applying the 
rule to this polynomial would be f(cc), since the succession of 
coefficients is the same for both polynomials up to a^+i- By 
the rule the final remainder is obtained by multiplying the expres- 
sion just obtained, in this case f(cc), by a and adding the last 
coefficient, in this case a^_^,^. This affords the final remainder 



170 



ADVANCED ALGEBRA 



EXERCISES 

1. Prove by complete induction that the partial remainders up to the final 
remainder obtained in the process of synthetic division are the coefficients of 
the quotient of f{x) hj z — a. 

2. Perform by synthetic division the following divisions. 

(a) a;3 - 7x2 _ 6x + 72 by a; - 4. 

Solution : 1-7-6 + 72 [4 

4 _ 12 - 72 



1 _ 8 - 18 
Quotient = a;2_3x- 18. 

(b) aj8 - 9x + 10 by X - 2. (c) 4x3 - 7 x - 87 by x - 3. 

(d) x3 + 8x2 - 4x - 32 i3y x-2. (e) x^ + 4x2 - 7x - 30 by x + 3. 

(f) x8 - 6x2 + 11 X - 6 by X - 1. 

(g) x* - 16x3 + 86x2 - 176x + 105 by x2 - 8x + 7. 

Hint. SinceK2— 8a;+7=(a; — 7)(x — 1), divide by x—7 and the quotient by a;— 1. 

(h) x6 + 1 by X + 1. (i) x9 - 1 by X - 1. 

(j) x4 + x3 - X - 1 by x2 - 1. (k) x^ - 2x3 - 4x by x - 3. 

(1) x6 - 2x8 - 4x - 1 by X + 2. (m) 4x3 - 6x2 - 2x - 1 by x - 3. 

(n) 2x* + 5x3 - 37x2 +44x + 84 by x2 + 6x - 6. 

163. Plotting of equations. We can now form the table of 
values necessary to plot an equation of the type 

^0^" + ^i^'"" ^ H 1- «„-i^ + «„ = y. 

Example. Plot x3 + 4 x2 - 4 = j/. 
l + 4 + 0-4[l 

+1+5+5 
1+6+6+1 
1 + 4 + 0-41-1 

-1-3+3 
l+S-S-l 

1 + 4 + 0-41-2 

-2-4+8 
1+2-4+4 
l_f.4 + 0-4[-_3 

-3-3+9 

+1-3+5 
1 + 4 + 0-41-4 

-4+0+0 



X 


V 


X 


y 





-4 


- 1 


-1 


1 


+ 1 


-2 


+ 4 


;l xo 


-3 


+ 6 






-4 


-4 



1+0+0-4 




THEORY OF EQUATIONS 171 

In this figure two squares are taken to represent one unit of x. A single 
square represents a unit of y. 

By an inspection of the figure it appears that the curve crosses the X axis 
at about x = .8, x = — 1.2, and x = — 3.7. Thus the equation for y = has 
approximately these values for roots (§ 110). 

164. Extent of the table of values. Since the object of plot- 
ting a curve is to obtain information regarding the roots of its 
equation, stretches of the curve beyond all crossings of the X axis 
are of no interest for the present purpose. Hence it is desirable 
to know when a table of values has been formed extensive enough 
to afford a plot which includes all the real roots. If for all values 
of X greater than a certain number the curve lies wholly above 
the axis, there are no real roots greater than that value of x. 

By inspection of the preceding example it appears that if for 
a given value of x the signs of the partial remainders are all 
positive, thus affording a positive value of ?/, any greater value 
of X will afford only positive partial remainders and hence only 
positive values of y. 

Thus when ■ all the partial rem.ainders are positive no greater 
positive value of x need he substituted. 

Similarly, when the partial remainders alternate in sign begin- 
ning with the coefficient of the highest power of x, no value of x, 
greater negatively, need be substituted. 

In plotting, if the table of values consists of values that are 
large or are so distributed that the plot would not be well propor- 
tioned if one space on the paper were taken for each unit, a scale 
should be so chosen that the plot will be of good proportion, 
that is, so that all the portions of the curve between the extreme 
roots shall appear on the paper, and the curvatures shall not be 
too abrupt to form a graceful curve. This was done, for example, 
in the figure, § 163. 

EXERCISES 

Plot and measure the values of the real roots of the equations when y = 0. 
1. x8 - 7 X - 6 = y. 2. x3 - 7 X + 5 = y. 

3. 7x8 - 9x - 6 = y. 4. x3 - 31 X + 19 = y. 

6. x3 - 12x - 14 = y. 6. 4x8 - 13x + 6 = y. 



172 abvakced algi:biia 

7. a;8 - 12x - 16 = y. 8. x^ - 45x + 152 = y. 

9. x* - 2x3 - X + 2 = y. 10. 8x3 _ igxS + 17x - 6 = y. 

11. X* - 17x2 + X + 20 = y. 12. x-i - 4x3 + 9x2 - 8x + 14 = y. 

13. 18x3-36x2 + 9x + 8 = 2/. 14. x* + 5x3 + 12x2 + 52x- 40 =y. 

15. x*-2x3-7x2+19x-10=y. 16. x4-6x3 + 3x2 + 26x-24 = y. 
17. 6x4 - 13x3 + 20x2 - 37x + 24 = y. 

165. Roots of an equation. In the case of the linear and 
quadratic equations we have been able to find an explicit value 
of the roots in terms of the coefficients. Such processes are prac- 
tically impossible in the case of most equations of higher degree. 
In fact the proof that any equation possesses a root lies beyond 
the scope of this book, and we make the 

Assumption. Every equation possesses at least one root. 

This is equivalent to the assumption that there is a number, 
rational, irrational, or complex, which satisfies any equation. 

166. Number of roots. We determine the exact number of 
roots by the following 

Theorem. Every equation of degree n has n roots. 

Given the equation /(x) = ao^?" + aiic""^ -\ +- a„ = 0. 

Let ^i (see assumption) be a root of this equation. Then (p. 166) 
ic — ^i is a factor of the left-hand member, and the quotient of 
f{x) by £c — «! is a polynomial of degree n —1. Suppose that 

aoic" -f- «ia;«- 1 -f ••• + «« = «o(^ - «^i) (ic"" ^ + ^i^c""^ H h ^„- 1). 

By our assumption the quotient a;"" ^ -f Jicc'*"^ 4- • • • + ^„_ i = 
has at least one robt, say az, to which corresponds the factor 
X — a^. Thus 

f{x) =aQ{x- a^ {x — a^) (a;"-^ -\- CiX"-^ -\ h c^.g). 

Proceeding in this way we find successive roots and corre- 
sponding linear factors until the polynomial is expressed as the 
product of n linear factors as follows : 

f{x) =ao(x- ai) (x-a2)'"(x- a„) = 0, 
where the roots are Uu ag, • • • , a„. 

Remark. This theorem gives no information regarding how many of the roots 
may be real or imaginary. This depends on the particular values of the coefl&cients. 



THEORY OF EQUATIONS 173 

Corollary. Any polynomial in x of degree n may he expressed 
as the product of n linear factors of the form x — a, where a is 
a real or a complex number. 

It should be noted that the roots are not necessarily distinct. 
Several of the roots and hence several of the factors may be 
identical. 

If f(x) is divisible by {x — aiY, that is, if a^ = a^, we say that 
oTi is a double root of the equation. Similarly, \i f(x) is divis- 
ible by {x — oTi)'', a^ is called a multiple root of order r. When 
we say an equation has n roots we include each multiple root 
counted a number of times equal to its order. 

Theorem. An equation of degree n has no more than n 
distinct roots. 

Let/(x) = ^0^" H f" ^n = ^ have the roots a-^^, cc2j'"j *».• Write 

the equation in the form 

ao(x — ai) • • ' (x — a^) = 0. 

If r is a root distinct from a^, •••,«:„, it must satisfy the equation 
and 

ao(r-aj)"'(r-a^) = 0. 

Since this numerical expression vanishes one of its factors 
must vanish (§5). But r =f= ai, thus r — a^ =^ 0. Similarly, no 
one of the binomial factors vanishes. Thus (§5) «<, — 0, which 
contradicts the hypothesis that the equation is of degree n. 

This theorem may also be stated as follows : 

Corollary I. If an equation a^x" + a^^x"" "^ + \- a^= of 

degree n is satisfied by more than n values of x, all its coefficients 
vanish. 

The proof of the theorem shows that if the equation has n + 1 
roots, ao = 0. We should then have remaining an equation of 
degree n —1, also satisfied hj n +1 values of x. Thus the coeffi- 
cient of its highest power in x vanishes. Similarly, each of the 
coefficients vanishes. 



174 ' ADVANCED ALGEBRA 

Corollary II. If two ^polynomials in one variable are equal 
to each other for every value of the variable^ the coefficients of 
like ^powers of the' variable are equal and conversely. 

Let a^x"" + a^x""-^ H h o^„ = ^o^" + ^i^c""^ H h *„ 

for every value of x. 

Transpose, {a^ — h^x''-\--'--\-a^ — h^ = ^. 

By Corollary I, a^ — h^^^ 0, or a^ = b^, 

ai — bi = 0, or a^ = bi, 

(^n-K = 0, or a„ = ^>„. 

167. Graphical interpretation. The graphical interpretation 
of the theorems of the preceding section is that the graph of an 
equation of degree n cannot cross the X axis more than n times. 
Since each crossing of the X axis corresponds to a real root, there 
will be less than n crossings if the equation has imaginary roots. 

168. Imaginary roots. We now show that imaginary roots 
occur in pairs. This we prove in the following 

Theorem. If a -\- ib is a root of an equation with real coeffi- 
cients, a — ib is also a root of the equation. 

li a -{- ib is a root of the equation <Xo^" + a^x^' ^ -\- -••-{- a^ = 0^ 
then X —(a -{- ib) is a factor (p. 166). We wish to prove that 
X — {a — ib) is also a factor, or what amounts to the same thing, 
that their product 

[x —{a-\- ib)'\ [x —{a — lb)'] = [(x — a) — ib"] [(x — a)-\- ib"] 

is a factor of f(x). Divide f(x) by (x — a)^ + b^ and we get 

f(x) = Q(x) [(x - af J^b^-]J^rx + r', (1) 

where r and r' are real numbers. This remainder rx + r' can be 
of no higher degree in x than the first, since the divisor 

(x - af + b'' 



THEORY OF EQUATIONS 



175 



is only of the second degree (§ 26). Now this equation (1) being 
an identity is true whatever value is substituted for x, as, for 
instance, the root of /(x), a -\- ib. Substituting this value for x, 
we get i 

f{a Jrib) = = Q{a + ib) \_{a + ib - ay + b^-]^^ r {a -\- ib) + r', 

or (p. 33) and (p. 3) = + r«^ -f r' + irb, 

or (p. 156) m + r' = 0, (2) 

rb = 0. (3) 

Since b ^ 0, by (3), p. 3, r = 0. 

Also from (2), r' = 0. > 

Hence rx -{- r' = 0. 

Consequently there is no remainder to the division of f(x) by 
(x — a)^ -\- U\ and hence if a + ib is a root oif(x), a — ib is also a 
root. 

Corollary. Every equation of odd degree with real coeffi- 
cients has at least one real root. 

The roots cannot all be imaginary, else the degree of the equa- 
tion would be even by the preceding theorem. 

169. Graphical interpretation of imaginary roots. When we 
plot the equations 



y = x^ + Ax''-4. (1), 
Yk 



y = x^-i-4:x^-l (2), 





176 



ADVANCED ALGEBRA 



7/ = ^^ + 4^2 (3), 






y 


= 


a;8 


+ 4 


x" 


+1 (4 


0. 








^ 


\ 














/ 






/ 




\ 














/ 






/ 






>v 












/ 






r 






\ 












/ 










^ 


\ 






















\ 
























\ 


























\ 
























\ 


























\ 


J 




























x^ 


, 





















— 


— 


X 



we see that corresponding to the increase of the constant term is 
a corresponding elevation" of the curve with respect to the X axis. 
In fact in each case the curve is the same, but the value of y is 
gradually increased. In (1) and (2) we have three real roots, in 
(3) the curve touches the X axis, and in (4) we have only one 
real root. As the elbow of the curve is raised and fails to intersect 
the X axis a pair of roots cease to be real, and since a cubic equa- 
tion always has three roots, a pair of roots become imaginary. 
Thus we have the 

Pkinciple. Corresponding to every elbow of the curve that 
does not intersect the X axis there is a pair of imaginary roots 
of the equation. 

The converse is not always true. It is not always possible to 
find as many elbows of the curve which do not meet the X axis 
as there are pairs of imaginary roots. 



EXERCISES 

Plot the following equations and determine from the plot how many roots 
are real. 

1. X* - 1 = y. 2. a;6 - 2 = y. 3. x^ - x - 1 = y. 

4. x4 + l = y. 5. x* + x + l = ?/. 6. x* + 2x2 + 2 = y. 

7. x8 - 3x2 - X + 1 = y. 8. x8 - 2x2 + 4x - 1 = y. 

9. 2x8 + 3x2 + 6x + 6 = y. 10. x^ - 3x2 - 4x - 5 = y. 



THEORY OF EQUATIONS 177 

170. Relation between roots and coefficients. If we write the 
expression (Corollary, p. 173) 

and multiply the factors, we obtain by equating coefficients of like 
powers of x (p. 174) relations between the roots and the coeffi- 
cients. Take for example n = S. 

x' + Wx^ + hx + b, = (x- 13,) (x - 13,) (X - p,) 

= x'- (13, -{-/3, + 13,) x' + ()8i/82 + ^2^3 + ft A) X - )8i A^3 = 0. 
. Hence j^ = _ (/3^ + ^^ + ^3), 

*2 = Aft + Aft + A A, 

^3 = -AAft- 

This suggests the 

Theorem. The coefficient of ^""^ is equal to the sum of the 
roots with their signs changed. 

The constant term is equal to the product of the roots with 
their signs changed. 

In general the coefficient of a?""*" is equal to the sum of all 
possible products of r of the roots with their signs changed. 

We prove this theorem by complete induction. 
First. We have already established the theorem for equations 
of degree two on p. 106 and for equations of degree three above. 
Second. Assume the theorem for n =^m. That is, if 

£c- + ^'1^'"-' + • • • + *«. = (^ - A) (^ - A) • • • (^ - A). (1) 
we assume that h^, the coefficient of ic'"-'", is the sum of all possi- 
ble products of r of the numbers — A? — A? ' • '? ~ A- 

Multiply both sides of (1) hy x — A« + i. Denote the result by 
^m+i + b^^x^ + . . . + 5'^^^ = (x- p,)(x ^p,)...(x- ^^ + 0- (2) 

The term in 0;"'+^-'" in this equation is obtained by multiplying 
the terms b.x'^-'- and &^_ia;'»+^-'' in (1) by x and - p^+i respec- 
tively. That is, in (2) 

b'r = K + b^-,(-(im^O' (3) 

Now all possible products of r of the quantities — A> —ft* 
.. •, — A 4.1 may be formed as follows : (1) Neglect — ft + i, and 
form all possible products of r of those remaining. The sum of 



178 ADVANCED ALGEBRA 

these is b^. (2) Form all possible products of r - 1 of — )8i 
— Aj • • •> — Pm^ not including — Pm + u and multiply each product 
^y — Pm+\- Add all the products obtained. This process, it is 
observed, is precisely that indicated by (3). 

Remark. It is noticed that in the rule the signs of the roots are always changed 
before forming any term. This does not involve any change when r is an even 
number, but is included in the rule for the sake of uniformity. 

Corollary. Every root of an equation is a factor of its con- 
stant term. 

171. The general term in the binomial expansion. On p. 129 

we gave an expression for the (r + l)st term of the binomial ex- 
pansion, the validity of which we now establish. In (1), § 170, 
let ySi = ^2 = • • • = Pn- Denote this common value by — a. The 
expression (1) becomes, on writing n in place of m, 

X- + hx""-' + --- + b,=(x + ay. 
By the theorem in § 170, b^ is the sum of all possible products 
of r of the negative roots. Since there are 



a.= 



n(n — 1) "• (n — r -i-l) 



ri 



such products, and since the roots are now identical, we obtain 
n(n-l)>-'(n-r-{-l) ^^_^^^ 
rl 
as the form of the (r + l)st term of the expansion of (x + a)". 

172. Solution by trial. Since by the previous corollary every 
root of an equation is a factor of its constant term, we may in 
many cases test by synthetic division whether or not a given equa- 
tion has integral roots. Thus the integral roots of the equation 

ic* _ 8a;» + 4:ic=» -f 24a; - 21 = (1) 

must be factors of 21. 

We try + 1 by synthetic division, ' 

1_8h-4 + 24- 2111 

-|.1_7_ 3-1-21 
1_7_3-|-21 
Thus 1 is a root of (1), and the quotient of the equation by 
"^-^^^ a;»-7x='-3a;-f 21 = 0. (2) 



THEORY OF EQUATIONS 179 

If this equation has any integral root it must be a factor of 21. 
We try + 3 by synthetic division, 

1_7_ 3 + 21|3 

+ 3 _ 12 - 45 
1-4-15-24 

Thus 3 is not a root. We try + 7, 

1_7_3 + 21|7 
+ 7 4- - 21 



1+0-3 
Thus 7 is a root, and the remaining roots of (1) are the roots of 
a;2 - 3 = 0, 
that is, X =± V3. 

Hence the roots of (1) are + 1, + 7, ± V3. 

EXERCISES 

Solve by trial : 

1. x3 - 7x2 + 50 = 0. 2, x3 - 9x 4- 28 = 0. 

3. x3-36x- 91 = 0. 4. x3 + 9x + 26=:0. 

5. x8 - 19x + 30 = 0. 6. x3 - 27x - 54 = 0. 

7. x8 + 2x2 - 23x + 6 = 0. 8. x3 - 6x2 + 11 X - 6 = 0. 

9. x3 - 2x2 - llx + 12 = 0. 10. x3 - 8x2 ^ 19a; _ 20 = 0. 

11. x8 + 9x2 + 27x + 26 = 0. 12. x* - 8x3 + 8x2 ^ 40x - 32 = 0. 

13. X* - 13x2 + 48x - 60 = 0. 14. x* - 3x3 - 34x2 + 18x +168 = 0. 

15. X* +8x3- 7x2 -50x + 48 = 0. 

16. x* - 3x3 - 5x2 + 29x - 30 = 0. 

17. x* - 6x3 + 13x2 - 30x + 40 = 0. 

18. x4-8x3+21x2-34x + 20 = 0. 

19. x* - 12x3 + 43x2 - 42x + 10 = 0. 

173. Properties of binomial surds. A binomial surd is a 

number of the form a ± V^, where a and b are rational numbers, 
and where b is positive but not a perfect square. 

Though we have not explicitly defined what we mean by the sum of an 
irrational number and a rational number, we shall assume that we can 
operate with the binomial surd just as we would be able to operate if b were 
a perfect square. 



180 ADVANCED ALGEBRA 

Theorem \. If a binomial surd a + V^ = 0^ then a =0 anc 
b = 0. 

J,i a-\- V^ = and either a = ov b = 0, clearly both must equal 
zero. Suppose, however, that neither a nor b equals zero. Then 
transposing we have a =^ 'Vb, and a rational number would be 
equal to an irrational number, which cannot be. Hence the only 
alternative is that both a and b equal zero. 

Theorem II. If two binomial surds, asa -\- -sib and c + V5, are 
equal, then a = c and b = d. 

Let a + V^ = c -f V^. 

Transposing c, a — c -\- V^ = V5. (1) 

Square and we obtain 

(^ci - cy-\-b -\- 2(a - c)Vb = dy 
or 

(^a, - cy + b - d -\- 2(a - c) Vb = 0. 

Thus, by Theorem I, either b = 0, which is contrary to the defi- 
nition of a binomial surd, or a ~ e = 0, that is, a = c. In the 
latter case (1) reduces to Vb == Vd, OTb = d, and we have a = c and 
b = dy which was to be proved. 

a 4- Vb and a — V^ are called conjugate binomial surds. 

Theorem III. If a given binomial surd a + Vb is the root of 
an equation with rational coefficients, then its conjugate is also a 
root of the same equation. 

The proof of this theorem, which should be performed in writ- 
ing by each student, may be made analogously to the proof of 
the theorem on p. 174. 

174. Formation of equations. If we know all the roots of an 
equation, we may form the equation in either one of two ways 
(see p. 167 and p. 177). 

First method. If a^, oc^,-", oc^ are the given roots, multiply 
together the factors x — a^,- ■•,x — a^. 

Second method. From the given roots form the coefficients 
by the rule on p. 177. 



THEORY OF EQUATIONS 181 

If the equation and all but one of its roots are known, that 
root can be found by the solution of a linear equation obtained 
from the coef&cient of the second or the last term. If all but two 
of its roots are known, the unknown roots may be found by the 
solution of a pair of simultaneous equations formed from the 
same coefficients. 

In the solution of the following exercises use is made of the 
theorem on p. 174, Theorem III, p. 180, and the various relations 
between the roots and the coefficients. 

EXERCISES 

1. Form the equations which have the following roots. Check the process 
by using both methods of § 174. 

(a) 2, - 3, 1. 
Solution : 

First metfiod. {x - 2){x + 3) (x - 1) = JC^ - 7 x + 6. 

Second method. Let the equation be 

x3 + 6ix2 + biX + 63 = 0. 
Then, by § 170, 61 = - (2 - 3 + 1) = 0, - • \ 

62 = -6 + 2-3 = -7, 

63 = -2-3. -1 = 6. 
The equation then is x^ — 7x + 6 = 0. 

(b) 1, 2, 3. (c) 2, 2, 2, 2. 
(d) 3, 1, 1, 0. (e) 1, 0, 0, 0. 
(f) ±V2,±i. (g) 2,4, -6. 
(h) 2, - 3, 1, 0. (i) 2, 3, - 6. 

. (j) 7, V5, -V5. (k) 1,2, -1, -|. 

(1) 3, 1 + i, 1 - i (m) _ 4, - 3, 3 ± V5. 

(n) 1 ± i, - 1 ± i (o) 2, V=^, -V^^. 

(p) -1,2,3,-4. (q)2^,3|, -H, -2^. 
(r) ±V6, ±iV7. (s) -5, 2 + V6, 2-V5. 

(V) 3, ^i±^, i:!^. (w) - 1, l±i^^ 1^. 



182 



ADVANCED ALGEBRA 



2. The equation x* + 2x3 - 7^2 _ 8x + 12 = has two roots — 3 and 
+ 1. Find the remaining roots. 

Solution : Let the unknown roots be a and h. 
Then, by § 170, _a-6 + 3-l = 2, 

- 3 a6 = 12. 
Solving for a and 5, we obtain a = — 2 or +2, 

6 = + 2 or - 2. 

3. x8 — 7x + 6 = has the roots 2 and 1. Find the remaining root. 

4. X* — 3x + 2 = has the root 1. Find the remaining roots. 

5. x^ — 18 X — 35 = has the root 5. Find the remaining roots. 

6. Two roots of x* — 35 x^ -f- 90 x — 56 = are 1 and 2. Find the remain- 
ing roots. 

7. The roots of x^ - 6 x2 - 4 x + 24 = are in A. P. Find them, 

8. The two equations x3-6x2 + llx-6 = and x^ -14x2 + 63 x - 90 = 
have a root common. Plot both equations on the same axes, and find all the 
roots of both equations. 

9. Determine the middle term of the equation whose roots are — 2, 
+ 1, 3, — 4 without determining any other term. 

10. What is the last term of the equation whose roots are — 4, 4, ± V— 3 ? 

11. One root of x* - 4x8 + 5x2 + 2x + 52 = is 3 - 2i. Find the remain- 
ing roots. 

12. One root of x* - 4 x^ + 5 x2 + 8 x - 14= is 2 + i V3. Find the others. 

13. Plot the following equations, determine all the integral roots, and 
find the remaining roots by solving. 

(a) X* - 6x8 + 24x - 16 = 0. 




X 


y 


-1 


- 33 


-2 





-3 


+ 155 



In this plot two squares on the X axis represent a unit of x, while one 
square on the Y axis represents ten units of y. The integral factors are 
X — 2 and x + 2, since ± 2 are roots, that is, are values of x for which the 



THEORY OF EQUATIONS 183 

curve is on the X axis. To find the quotient of our equation we first divide 
synthetically by 2, and then the quotient by — 2, using the principle given 
in § 161. 

1-6+ + 24- 16 12 
+ 2 - 8-16 + 16 
^ 1-4- 8+ 8 |-2 

- 2 + 12 - 8 
1-6+ 4 

Thus the quotient of the polynomial and (x — 2) (« + 2) is «2 __ ga; 4. 4 
Solving the equation 

x2-6x + 4 = 0, 

we obtain the two remaining roots, x = 3 ± V5. These remaining roots 
might also be found by the method of exercise 2. 

(b) x3 - 6x - 12 = 0. (c) x3 - 8x2 + 7 = 0. 

(d) x8 - 7x2 + 50 = 0. (e) x^ - 8x2 + 13x- 6 = 0. 

(f) x8-6x2+7x-2 = 0. (g) x3 + 3x2 + 4x- 24 = 0. 

(h) X* - 3x3 + 7 x2 - 21 X = 0. (i) X* - 3x3 - 7 x2 + 27 x - 18 = 0. 

'(j) x*-9x3 + 21x2-19x + 6 = 0. 

(k) How many imaginary roots can an equation of the 5th degree have ? 
(1) x3 — ax2 + 6x + c = has two roots whose sum is zero. What is the 
third root ? What are the two roots whose sum is zero ? 

(m) x3 + x2 + 6x + c = has one root the reciprocal of the other. What 
are the values of the roots ? 

(n) x3 — 4x2 + ox + 62 = has the sum of two roots equal to zero. What 
must be the values of a and h ? 

(o) X* - 3 x3 + 6x + 9 = has the sum of three of its roots equal to zero. 
What must be the value of 6 ? 

175. To multiply the roots by a constant. Suppose we have 
the equation 

fix) = a,x- +- a^x--' + . . . + «^ = 0, (1) 

whose roots are «ri, a^, • • • , «:„. An equation of this type for 
values of n greater than 2 is usually not solvable by elementary 
methods. It often happens, however, that by changing its form 
slightly we may obtain an equation one or more of whose roots 
we can find. We shall see that if an equation has rational roots 
we may always find them if we change the form of the equation 
as indicated on the following page. 
/ 




184 ADVANCED ALGEBRA 

We seek to form from (1) an equation whose roots are equal 
to the roots of (1) multiplied by a constant factor, as k. Thus 
the equation we seek must have the roots kai, ka^, ka^. We 
carry out the proof, which is perfectly general, on the equation 
of the third order 

f(x) = a^x^ -f- a-^x^ -f «2^ + ^3 = 0, 

whose roots are a^, a^, a^. The equation that we seek must have 
roots kai, ka^, ka^. Since now (§ 53) f{x) = is satisfied by 
a, where a stands for any one of the. roots, that is, since f(a) = 0, 

evidently /(t) = is satisfied by ka, that is. 

Hence we obtain an equation that is satisfied by ka^, ka^^ ka^, 
if in/(£c) we let a; = t* 

The required equation is then 

Akj-'W^l^^-k^''^-^^ 

or, multiplying by k^, 

a^z^ + ka-^z'^ -{- k^azZ -{- k^a^ — 0. 
This affords the general 

KuLE. To multiply the roots of an equation hy a constant k, 
multiply the successive coefficients beginning with the coefficient of 
x""'^ hy k,l^, • • -,1^ respectively. 

In performing this operation the lacking powers of x should be 
supplied with zero coefficients. 

Example. Multiply the roots of 2 x^ — 3 x + 4 = by 2. 
Multiply the coefficients by the rule above, 

2«8 + 2-0x2-4-3x + 8-4 = 0. 
Simplifying, »» - 6 x + 16 = 0. 



THEORY OF EQUATIONS 185 

When an equation in form (2), p. 166, has fractional coefficients, 
an equation may be formed whose roots are a properly chosen 
multiple of the roots of the original equation and whose coeffi- 
cients are integers. 

Corollary I. When k is a fraction this method serves to 
divide the roots of an equation hy a given number. 

Corollary IL When k = — 1 this method serves to form an 
equation whose roots are equal to the roots of the original equa- 
tion hut opposite in sign. This is equivalent to the statement 
that /(— x) = has roots equal hut opposite in sign to those 

• EXERCISES 

1. Form the equation whose roots are three times the roots of 

Solution : Supplying the missing term in the equation, we have 

«* -6x3 + 0x2- X + 1 = 0. 
Since A; = 3, we liave by the rule 

x* -3 -6x3 + 9 -0x2- 27 -x + 81 = 0, 'r 

or x*-18x3 -27x + 81 = 0. ' 

2. Find the equation whose roots are twice the roots of 

x* + 3x3-2x + 4 = 0, 

3. Find the equation whose roots are one half the roots of 

x3-2x2 + 3x-4 = 0. 

4. Find the equation whose roots are two thirds the roots of 

x3-4x-6 =0. 

5. By what may the roots of the following equations be multiplied so thai 
in the resulting equation the coefficient of the highest power of x is unity 
and the remaining coefficients are integers ? Form the equations. 

(a) 3x3- 6x + 2 = 0. 

Solution : We wish to bring into every term such a factor that all the 
resulting coefficients are divisible by 3. 

Let A: = 3. 

Supply the lacking term, 

3x3 + 0x2 -6x + 2 = 0. 

By rule, 3x3 + 3 • 0x2 - 9 • 6x + 27 • 2 = 0. 

• Dividing by 3, x3 - 18 x + 18 = 0. 



186 



ADVANCED ALGEBRA 



x2 

(b) X3 + - - 



1 = 0. 



(C) X8 - i = 0. 



(d) «» + 7x2 + -x + - = 0, 



62 



6» 



(e) x4 + ^ 



x2 



+ 1 = 0. 



(f) 2x8 - 3x2 -x + 4 = 0. 
(h) x*-6x3-2x2+ 1 =0. 



(g) 3x4-3x2-4x + l = 0. 
(i) 16x4 - 24x3 + 8x2 -2x + 1 = 0. 



6. Form equations whose roots are the negatives of the roots of the fol- 
lowing equations. 

(a) x8-4x + 6 = 0. 

Solution : Supply the lacking term, 

x3+ 0x2-4x + 6 = 0. 
Changing signs we obtain by Corollary II 

x3_0x2-4x- 6 = 0, 
or x3 - 4 X - 6 = 0. 

(b) X' - 2x2 - 4x = 0. (c) x* - 3 x2 + 1 = 0. 

(d) X* - 2x3 + x2 4- 2x - 1 = 0. (e) x3 + 3x2 + 7x - 13 = 0. 

7. What effect does changing the sign of every term of the member 
involving x have on the graph of an equation? 

8. What is the graphical interpretation of the transformation which 
changes the signs of the roots of an equation, that is, what relation does 
the graph of the equation before transformation bear to the graph of the 
equation after transformation (a) when the degree is an even number, 
(b) when the degree is an odd number? 

9. If 4x4 - 16x3 - 86x2 + 4x + 21 = has as two roots - ^ and - 3, 
what are the roots of 4x4 + 16x3 - 85x2 - 4x + 21 = 0? 

10. If a root of xs - 11 x2 + 36 x - 36 = is 2, what are the roots of 
x»+ 11x2+ 36x + 36 = 0? 



176. Descartes' rule of signs. A pair of successive like signs in 
an equation is called a continuation of sign. A pair of successive 
unlike signs is called a change of sign. 



In the equation 



2x4_3a;8 + 2x2 + 2a;-3 = 



(1) 



are one continuation of sign and three changes of sign. This may be seen more 
clearly by writing merely the signs, + — + + —. 

Let US now inquire what effect if any is noted on the number 
of changes of sign in an equation if the equation is multiplied by 



THEORY OF EQUATIONS 187 

a factor of the form x — a when a is positive, that is, when the 
number of positive roots of the equation is increased by one. 
Let us multiply equation (1) by a; — 2. We have then 

x-2 

2x^ -3x*-{-2x^-{-2x''-Sx 

-4:X^+6x^-4:X^-4:X-\-6 

2x' - 7 x^ -{- Sx^ - 2x^ - 7 x -{- 6 

In this expression the succession of signs is + — + — — +, 
in which there are four changes of sign, that is, one more change 
of sign than in (1). If an increase in the number of positive roots 
always brings about at least an equal increase in the number of 
changes of sign, there must be at least as many changes of sign in 
an equation as there are positive roots. This is the fact, as we 
now prove. 

Descartes' rule of signs. An equation f(x) = has no more 
real positive roots than f(x) has changes of sign. 

Illustration. In the equation of degree one £c — 2 = there 
is one change of sign and one jjositive root. In the case of a linear 
equation there is no possibility of more than one change of sign. 
In the quadratic equation x'^-\-2x-\-l = Q there is no change of 
sign, and also no positive root since for positive values of x the 
expression x^ -f 2 cc + 1 is always positive and hence never zero. 
In the equation ic^ + 2ic — 3 = we have one change of sign, 
and one positive root, + 1. 

We shall prove this general rule by complete induction. 

First. W^e have just seen that the rule holds for an equation 
of degree one. 

Second. We assume that the rule holds for an equation of 
degree m, and prove that its validity for an equation of degree 
m -\-l follows. We shall show that if we multiply an equation 
of degree m hj x — a, where a is positive, thus forming an equa- 
tion of degree m + 1, the number of changes of sign in the new 
equation always exceeds the number of changes of sign in the 



188 



ADVANCED ALGEBRA 



original equation by at least one. That is, the number of chanj 
of sign increases at least as rapidly as the increase in the number 
of positive roots when such a multiplication is made. 

Let/(£c) = represent any particular equation of the nth. degree. 
The first sign of f(x) is always +. The remaining signs occur in 
successive groups of + or — signs which may contain only one 
sign each. If any term is lacking, its sign is taken to be the same 
as an adjacent sign. Thus the most general way in which the 
signs of /(x) may occur is represented in the following table, 
in which the dots represent an indefinite number of signs. The 
multiplication of f(x) by a; — a is represented schematically, onl^P"-^ 
the signs being given. ^ 



fix) 

X — a 


All + signs 
+ •••• + 


All - signs 


All + signs 


All - signs 


Further 
groups 

+ •••• + 


All - signs 

+ - 


xf{x) 
- ocfix) 


+ + ••• + 


- + ••• + 


+ + ••• + 
+ 


- + ••• + 


+ +••• + 
+ 


-+ + + 


{x-a)f{x) 


+ d=---± 


-±.-.± 


+ ±---± 


-±...± 


+ ± ± 


-±•••±4- 



The ± sign indicates that either the + or the — sign may occur 
according to the value of the coefficients and of a. The verti- 
cal lines denote where changes of sign occur in f(x). Assuming 
that all the ambiguous signs are taken so as to afford the least 
possible number of changes of sign, even then in (x — cc)f(x) 
there is a change of sign at each or between each pair of the 
vertical lines, and in addition, one to the right of all the vertical 
lines. Thus as we increase the number of positive roots by one 
the number of changes of sign increases at least by one, perhaps 
by more. 

The only possible variation that could occur in the succession 
of groups of signs in f(x), namely, when the last group is a 
group of 4- signs, does not alter the validity of the theorem. 



THEORY OF EQUATIONS 189 

We illustrate the foregoing proof by the following particular 
example. 

Let f{x) = x^ -4:X^ - x-{-2, and let a = 2. 

Multiply f(x), 1 + 0-4 + 0-1 + 2 4 changes 

by x-2, 1-2 

xf(x), 1 + 0-4 + 0-1 + 2 

-2f(x), -2-0 + 8-0 + 2-4 

(x~2)f(x), 1-2-4 + 8-1 + 4-4 5 changes 

177. Negative roots. Since /(—a:) has roots opposite in sign 
to those of f(x) (p. 185), we can state 

Descartes' rule of signs for negative roots. f(x) has nc 
more negative roots than there are changes in sign in /(— x). 

If by Descartes' rule it appears that there cannot be more than 
a positive roots and b negative roots, and if a -{- b <n, the degree 
of the equation, then there must be imaginary roots, at least 
n — (a -{- b) in number. 

EXERCISES 

1. Prove Descartes' rule of signs for x^ + bx + c = directly from the 
expression for b and c in terms of the roots (see § 115). 

2. Find the maximum number of positive and negative roots and any 
possible information about imaginary roots in the following equations. 

(a) x3 + 2 x2 + 1 = 0. 

Solution : Writing signs of f{x), + + +, there is no change, hence no 
positive root. 

Writing signs of /(— £c), — + + , there is one change, hence no more 
than one negative root. Since there can be only one real root there must 
be two imaginary roots. 

(b) x3 + 1 = 0. (c) X* - 2 = 0. 

(d) x3 - X + 1 = 0. (e) x6 - X + 1 = 0. 

(f ) x4 + X + 1 = 0. (g) x6 + x2 + 1 = 0. 

(h) x3-6x2+4x-l = 0. (i) x5-2x4-3x3+4x2+x + l = 0. 

(i) x6 + 2x* - 6x3 - 4x2 + X - 1 = 0. 



190 ADVANCED ALGEBRA 

178. Integral roots. In finding the rational roots of an equa- 
tion we make use of the following 

Theorem. If the equation 

^ + «i^"-'+---+a„ = ^ (1) 

{where the oUs are integers) has any rational root, such root must 
he an integer. 

Suppose - be a fraction reduced to its lowest terms which 
satisfies the equation. 

Then ^ + 2l£r! + ... + „„ = 

is an identity. 

Then clearing of fractions and transposing, 

f = --q {ciip"~^ H h a„!?"~0- 

Thus some factor of g' is a factor of ^^ that is, oip (p. 52), which 
contradicts the hypothesis that - is reduced to its lowest terms. 

Thus all the rational roots of the equation are integers, which 
as we know (§ 170) are factors of a„. 

179. Rational roots. If we seek the rational roots of 

aox"" H [- a„ = 0, 

where a^ =^ 1, we can multiply the roots by a properly chosen 
constant (§ 175) and obtain an equation of form (1) above whose 
integral roots may easily be found by synthetic division. 

Example. What rational roots, if any, has 

8x»+lla;-14 = 0? (1) 

Multiply the roots by 8, 3 a^ + 99 x - 378 = 0. 

Divide by 3, x^ + 33 x - 126 = 0. (2) 

Since by Descartes' rule of signs equation (1) has only one positive root 
and no negative root, we do not need to carry the table further than to test 
for a positive root. 



THEOKY OF EQUATIONS 191 

y Form a table of values for equation (2) by synthetic division. 

~^ We need only to try the factors of 126 (§ 170). 
_ g2 Thus (2) has the root 3. Hence the original equation (1) has 

Q the root 3^3 = 1. 



KuLE. To find all the rational roots of an equation, trans- 
form the equation so that the first coefficient is + 1. 

Find the maxiymum number of positive and negative roots by 
Descartes' rule of signs. 

Find the integral roots of this equation by trial, and the 
roots of the. original equation by dividing the integral roots 
found by the constant by which the roots were multiplied. 

By the Theorem § 178 we are assured that all the rational 
roots can be found in this way. 

EXERCISES 

Find all the rational roots of the following equations. 

1. 4x8 = 27(x + l). 2. 15jc8 + 13a;2-2 = 0. 

3. 4a;3-6x-6 = 0. 4. x^- 2f x2 + 2|x - 1 = 0. 

5. 4x3 - 8x2 - X + 2 = 0. 6. 3x* - 8x8 _ ^q^2 + 25 = 0. 

7. 4x8-4x2+ x-6 = 0. 8. 3x8+ 13x2+ llx - 14 = 0. 

9. 4x8 + 16x2 -9x- 36 = 0. 10. 2x3 - 21x2 + 74x - 85 = 0. 

11. 6x8 - 47x2 + 71X + 70 = 0. 12. 12x8 - 52x2+ 23x + 42 = 0. 

13. 6x8-29x2+ 53x- 45 = 0. 14. 6x*- x8- 8x2- 14x + 12 = 0. 
15. 27x3+ 63x2+ 30x- 8 = 0. 16. 2x*-13x8 + 16x2- 9x +20 = 0. 
17. 3x8 - 26x2 +52x- 24 = 0. 18. Ox* - x3 - 49x2+ 55x - 50 = 0. 
19. 18x8+ 81x2+ 121x + 60 = 0. 20. 12x4 + 6x8 - 24x2- 9x + 9 = 0. 

21. 10x4 + 18x8 - 16x2 + 8x- 20 = 0. 

22. 9x4 + 15x8 - 143x2 +41X + 30 = 0. 

23. 36x4 - 72x8 - 31 x2 + 67 x + 30 = 0. 

24. 24x4- 108x8+ 324x2 -240X + 60 = 0. 

180. Diminishing the roots of an equation. In the preceding 
sections we have solved completely the problem of finding the 
rational roots of an equation. We now pass to the problem of 



V 



192 ADVANCED ALGEBRA 



finding the approximate values of the irrational roots of an eqnar 
tion. In carrying out the process that we shall develop it is 
desirable to form an equation whose roots are equal respectively 
to the roots of the original equation each diminished by a constant. 

Let fix) = aox"" + a^x''-'^ -\ h «« = 0, (1) 

whose roots are ^i, a^, • ••, a^. Let a be any constant. We seek 
an equation whose roots are a^ — a, a^ — a, - • • , a^ — a. ^~^^^v_ 

If we let a stand for any one of the roots of (1), since f(a) = 
(p. 33), we see that 

/(« + a) = is satisfied by a — a, 
that is, f{a-a-\-a)= f(a) = 0. 

Thus to form the desired equation replace x by z -\- a. We 
obtain 

f(x) = f(z + a) = ao(z + a)» ■}- a,{z -{- af-^ + . . . + a„ = 0. 

Developing each term by the binomial theorem and collecting 
like powers of z, we get an equation of the form 

fix) = Fiz) = A,z- + A^z-^ + . . . + ^^ = 0, (2) 

where the ^'s involve the a's and d. This is the equation desired. 
We now seek a convenient method of finding the values of the 
coefficients ^o? ^i? ^2> •••> ^n when a^^ %, a^^ •••, a„ are given 
numerically. Now A^ is the remainder from the division of Fiz) 
by z. But since Fiz) =fix) and z = x — a, the remainder from 
dividing Fiz) by z is identical with' the remainder from dividing 
fix) hj x — a. Thus A^ is the remainder from dividing fix) by 
X — a. Furthermore, since A^_y^ is the remainder from dividing 

—^^ by z, it is also the remainder from dividing ^^-^^ 

by a; — a. The process may be continued for finding the other ^'s. 
We may then diminish the roots of an equation by a as follows : 

EuLE. The constant term of the new equation is the remaivr- 
der from dividing fix) hy x — a. 



i 



THEORY OF EQUATIONS 



193 



The coefficient of z in the new equation is the remainder from 
dividing the quotient just obtained hy x — a. 

The coefficients of the higher powers of z are the remainders 
from dividing the successive quotients obtained hy x — a. 

Example. Form the equation whose roots are 2 less than the roots of 
x4 _ 2 x3 - 4 x2 4- x - 1 = 0. 

The divisions required by the rule we carry out synthetically (p. 169). 
1-2-4+1- 1[2 
+ 2 +0 -8 -14 
-15 



1 + 

+ 2 


-4 -7 

+ 4 +0 


1 + 2 +0 
+ 2 +8 


-7 


1 + 4 

+ 2 


+ 8 





1+6 

The desired equation is 

^4 + 6x3 + 8x2 



7x-15 = 0. 



181. Graphical interpretation of decreasing roots. If an equa- 
tion has roots a units less than those of another equation, if a is 
positive its intersections with the X axis or with any line parallel 
to the X axis are a units to the left of the corresponding inter- 
sections of the first equation. It is, in fact, the same curve, except- 
ing that the Y axis is moved a units to the right. If a is negative, 
the Y axis is moved to the left. 



EXERCISES 

Plot, decrease the roots by a units, and plot the new axes. 

1. x4-3x3 -2x-3 = 0. a = 3. 

Solution: 1-3- 0- 2-3[3 



(1) 



+ 3-0 


- 


-6 


1_0 - - 2 
+ 3 + 9+27 


-9 


1 + 3 + 9 

+ 3 + 18 


+ 25 





1 + 6 

+ 3 

1 + 9 



+ 27 



194 



ADVANCED ALGEBRA 




Thus the required equation is 



0. (2) 



X 


y 





- 3 


: 1 


- 7 


2 


-16 


3 


- 9 


-1 


+ 3 



3. x3 _ 8 = 0. a = 1.4. 
5. x3 + 4x-8 = 0. a = 3. 
7. x3 + 2 X + 6 = 0. a = -\. 
9. «8-2x2 + 8x-7 = 0. a = 2. 

11. x4-3x2 + 2x 

12. X 

13. 2x3-6x2 + 4x -3 = 

14. X* + 6x3 + 10x2 + x - 1 = 



In the figure one square on the Faxis repre- 
sents two units of y, and two squares on the 
X axis represent one unit of x. 

2. x*-16 = 0. a = 2. 
4. x4- 2x2 + 1 = 0. a = .2. 
6. x3-4x2-2 = 0. a = .5. 
8. x3 + 4 x2 + X - 6 = 0. a = - .4. 
10. x3-3x2 + x-l = 0. a = -.3. 
2 = 0. a = -2. 
15x2 + 7x + 125 = 0. a = 5. - 

a = -3. 

a = -l. 



182. Location principle. If when plotting an equation y =f(x) 
the value x = a gives the corresponding value of y positive and 
equal to c, while the value x = b gives the corresponding value 
of y negative, say equal to — d, then the 
point on the curve x = a, y = c is above 
the X axis, and the point on the curve 
x = byy = — -^ is below the X axis. If our 
curve is unbroken, '^ must then cross the 
X axis at least once between the values 
X — a and x = h^ and hence the equation 
must have a root between those values 
of x. The shorter we can determine this 
interval a to b the more accurately we can find the root of the 
equation. This property of unbrokenness or continuity of the 
graph of ?/ = ^0^" 4- aix"~^ + • • • + a„ we assume. We assume 
then the following 




THEORY OF EQUATIONS 



195 



Location principle. When for two real unequal values of x, 
x = a and x = h, the value of y —f{x) has opposite signs, 
the equation f{x) = has a real root between a and h. 

Illustration. The equation f(x) =ic^-f3cc — 5 = has a 
root between 1 and 2. Since /(I) = - 1, /(2) = 9. 

183. Approximate calculation of roots by Horner's method. 

We are now in a position to compute to any required degree of 

accuracy the real roots of an equation. Consider for example 

the equation ^s + 3,_20 = 0. (1) 

Form the table of values for plotting the equation 

ic8 + 3 a: - 20 = ?/. 

By the location theorem we find 
that a root is between + 2 and + 3. 
To find more precisely the position 
of the root we might estimate from 
the graph the position of the root 
and substitute say 2.3, 2.4, and so 3+16 

on, until we found two values be- 
tween which the root lies. We can 
gain the same result with much _ i _ 24 
less computation if we first dimin- 
ish the roots of the equation so that 
the origin is at the less of the two integral values between which 
we know the root lies. Here we decrease the roots of (1) by 2, 

1 4- + 3 - 20[2 



y 
-20 

-16 




+ 2 + 4+14 


1 + 2 + 7 
+ 2 -h 8 


- 6 


1 + 4 
+ 2 


+ 15 





1+6 

The equation whose roots are decreased by 2 is 
a:» + 6ic2 + 15a;^6 = 0. 



(2) 



196 ADVANCED ALGEBRA 

We know that (2) has a root between and 1, since equation (1) 
has a root between 2 and 3. From the graph we can estimate the 
position of the root. Having made an estimate, say .3, it is neces- 
sary to verify the estimate and determine by synthetic divi^io^ 
precisely between which tenths the root lies. Thus, trying .3, we" 

^^*^^^ 1 + 6.0 + 15.00 - 6.000[_^ 

+ 0.3+ 1.89 + 5.067 
1 + 6.3 + 16.89 - 0.933 () 

which shows that for x = .3 the curve is below the X axis, hence 
the root is greater than .3. But we are not justified in assuming 
that the root is between .3 and .4 until we have substituted .4 
for X. This we proceed to do. 

1 + 6.0 + 15.00- 6.000 [^ 

+ 0.4+ 2.56 + 7.024 
1 + 6.4 + 17.56 + 1.024 

Since the value of y is positive for x = .4, the location principle 
shows that (2) has a root between .3 and .4, that is, (1) has a root 
between 2.3 and 2.4. 

To find the root correct to two decimal places, move the origin up 
to the lesser of the two numbers between which the root is now 
known to lie. The new equation will have a root between and .1. 

This process is performed as follows : 

1 + 6.0 + 15.00 - 6.000[^- 
+ 0.3 + 1.89 + 5.067 



1 + 6.3 + 16.89 
+ 0.3 + 1.98 



1 + 6.6 
+ 0.3 



0.933 



+ 18.87 



1 + 6.9 

Thus the new equation is 

x^ + 6.9 x^ + 18.87 X f .933 = 0. (3) 

This equation has a root between and .1. We can find an 
approximate value of the hundredths place of the root by solving 



THEORY OF EQUATIONS 197 

the linear equation 18.87 x — .933 = 0, obtained from (3) by drop- 
ping all but the term in x and the constant term. 

This suggestion must be verified by synthetic division to deter- 
mine between what hundredths a root of (3) actually lies. 

1 + 6.90 + 18.870 - 0.9330|.04 
+ 0.04 + 0.277 + 0.7658 

6.94 + 19.147 - 0.1672 

Thus the curve is below the X axis at a; = .04 and hence the 
root is greater than .04. We must not assume that the root is 
between .04 and .05 without determining that the curve is above 
the X axis at x = .05. 

1 + 6.90 -f 18.870 - 0.9330[^ 
+ 0.05+ 0.347 + 0.9608 

6.95 + 19.217 + 0.0278 

Thus the curve is above the X axis ai x = .05. By the location 
principle (3) has a root between .04 and .05, that is, (1) has a 
root between 2.34 and 2.35. We say that the root 2.34 is correct 
to two decimal places. If a greater degree of precision is desired, 
the process may be continued and the root found correct to any 
required number of decimal places. 

The foregoing process affords the following 

EuLE. Plot the equation. Apply the location principle to deter- 
mine between what consecutive positive integral values a root lies. 

Decrease the roots of the equation hy the lesser of the two 
integral values between which the root lies. 

Estinfiate from the plot the nearest tenth to which the root of 
the new equation lies, and determine hy synthetic division pre- 
cisely the successive tenths between which the root lies. 

Decrease the roots of this equation by the lesser of the two tenths 
between ivhich the root lies, and estimate the root to the nearest 
hundredth by solving the last two terms as a linear equation. 



198 



ADVANCED ALGEBRA 



Determine hy synthetic division precisely the hundredths inter- 
val in which the root must lie. ^ — ^ 

Proceed similarly to find the root correct to as many 'places ds 
may he desired. 

The sum of the integral, tenths, and hundredths values next less 
than the root in the various processes is the approximate value 
of the root. 

To find the negative roots of an equation f(x) = 0, find the 
positive roots of /(— x) = and change their signs. 

When all the roots are real a check to the accuracy of the com- 
putation may be found by adding the roots together. The result 
should be the coefficient of the second term. 



EXERCISES 

Eind the values of the real roots of the following equations correct to 
three decimal places. 

1, x3 + 4x2 + aj + l = 0. (1) 

Solution : Since by Descartes' 
rule of signs there are no posi- 
tive roots, we form the equation 
/( — a;) = and seek its positive 
root. 

Thus 
a;3-4x2 + aj-l = 0. (2) 

Plot the equation (2) set equal 
to y. In the figure two squares 
are taken as the unit of x. 

There is a root of this equa- 
tion between 3 and 4. 





Yj 


< 
















( 










































X 













































i ^ 




/ 




\ 














1 








> 














2 










k 








,f 


3 










\ 












4 












V 






J 
















\ 






/ 


















^ 


> 































y 
-1 

-3 

-7 
-7 
+ 3 



Decrease the roots of (2) by 3, 



1-4+1 
+ 3-3 



1[3 
6 



1-1 -2 

+ 3 +6 



1 + 2 

+ 3 

1 + 5 



+ 4 



THEORY OF EQUATIONS 199 

The equation is x^ + 5x^ + Ax — 7 = 0. (3) 

From the plot we estimate the root of (3) at .8. 

Verify, 1 + 5.0 + 4.00 - 7.000 [^ 

+ 0.8 + 4.64 + 6.912 
+ 5.8 + 8.64 - 0.088 

1+ 5.0 + 4.00- 7.000 [^ 
+ 0.9 + 5.31 + 8.379 
+ 5.9+9.31 + 1.379 

Thus the root is determined between .8 and .9. \ 
Decrease the roots of (3) by .8, 

1+5.0 + 4.00 -7.000^8 
+ 0.8 + 4.64 + 6.912 



1+5.8 + 8.64 
+ 0.8 + 5.28 



.088 



1+6.6 
+ 0.8 



+ 13.92 



1+7.4 



The equation is x^ + 7.4 x^ + 13.92 x - .088 = 0. (4) 

088 

Estimate the root of (4) at x = ~ = .006. 

^ ' 13.92 

Verify, 1 + 7.400 + 13.920000 - .088000 [.006 

+ 0.006 + 00.044436 + .083784 
+ 7.406 + 13.964 - .004216 
1 + 7 .400 + 13.920000 - .088000 [.007 
+0 .007 + 00.051849 + .097804 
+ 7.407 + 13.972 + .009804 
Thus the root of the equation (1) correct to three decimal places is - 3.80(> 
2. x3 - 4 = 0. 3. x* - 3 = 0. 

4. X3 + X = 20. 5. 3X4 - 5X3 zz: 3I. 

6. x3 + x2 = 100. 7. x3 - X - 33 = 0. 

8. x4 + X - 100 = 0. 9. x3 - 8x - 24 = 0. 

10. x4 - 4x3 + 12 = 0. 11. X* + x2 + X = 111. 

12. x3 - x2 + X - 44 = 0. 13. x3 + lOx - 13 = 0. 

14. x8 + 3 x2 - 2 X - 1 = 0. 15. x3 + x2 + X - 99 = 0. 

16. x3 - 9x2 - 2 X + 101 = 0. 17. X* + x3 + x2 - 88 = 0. 

18. X* - 12x3 - 16x + 41 = 0. 19. x3 - 6x2 + 5x + 11 = q. 

20. x3 - 10x2 + 35x+ 50 = 0. 21. 2x4-4x3 + 3x2-1 = 0. 

22. 3x4 - 2x3 - 21x2 -4x + 11 = 0. 23. 9x3 - 45x2 + 34x + 37 = 0. 



200 



ADVANCED ALGEBRA 



184. Roots nearly equal. Suppose we wish to find the positije 
roots, if any exist, of ) 

x^-{-17x^-A6x-i-29 = 0. (1) 



y 

4-29 

+ 1 

+ 13 
+ 71 



By Descartes* rule of signs we see that there can be 
only two positive roots. We obtain the adjacent table 
of values. From the plot that these values indicate we 
cannot tell whether any real root exists between 1 and 2, 
but if it does exist the plot indicates that it is nearer 
1 than 2. 

Decrease the roots of (1) by 1, 

1 + 17 -46 +29[1 
+ 1+18-28 



1 + 18 - 28 


+ 1 


+ 1+19 




1 + 19 


- 9 




+ 1 




1+20 




The new equation is 


x^-\-20x^- 


-9x- 


f 1 = 



0. 



(2) 



Estimate the root of (2) at .2 and carry the origin up to .2 



and also up to .3. 

1 + 20.0 - 9.00 + l.OOOj^ 
+ 0.2 + 4.04 - 0.992 



1 4- 20.0 - 9.00 + 1.000[J 
+ 0.3 + 6.09 - 0.873 



1 + 20.2 -4.96 
+ 0.2 + 4.08 



1 + 20.4 
+ 0.2 



+ 0.008 



1 + 20.3 - 2.91 
+ 0.3 + 6.18 



0.88 



1 + 20.6 
+ 0.3 



+ 0.127 



+ 3.27 



1 + 20.6 



1 + 20.9 



By Descartes' rule of signs on the numbers obtained by moving 
the origin to .3, it is seen that there are no positive roots of (2) 
greater than .3, while the rule would indicate that there might 
be roots greater than .2. We consider the equation 

ic^ + 20.6 x^ - 0.88 X + 0.008 = 0. (3) 



THEORY OF EQUATIONS 201 



Estimate the roots of (3) at 
_ 0.008 
^ ~ 0.88 



= .009.=* 



Verify, 1 + 20.60 - 0.880 -h 0.00800[.01 

+ 0.01 + 0.206-0.00674 
1 -f 20.61 - 0.674 + 0.00126 
1 + 20.60 - 0.880 + 0.00800|.02 

+ 0.02 + 0.412 - 0.00936 
1 + 20.62 - 0.468 - 0.00136 
This determines a root of (3) between .01 and .02. 

1 + 20.60 - 0.880 4- 0.00800|.03 

+ 0.03 + 0.619-0.00483 
1 + 20.63 - 0.161 + 0.00317 

This determines another root of (3) between .02 and .03. 
Decrease the roots of (3) by .01, 

1 + 20.60 - 0.880 + 0.00800|.01 
+ 0.01 + 0.206 - 0.00674 



1 + 20.61 - 0.674 
+ 0.01 +0.206 



+ 0.00126 



1 + 20.62 
+ 0.01 



- 0.468 



1 + 20.63 

The new equation is 

x^ + 20.63 x^ - 0.468 x + 0.00126 = 0. (4) 

Estimate the root of (4) at ic = ' ,^^ = .002.=* 
^ ^ 0.468 

Verify, 1 + 20.630 - 0.468 + 0.00126|.002 

+ 0.002 + 0.041 - 0.00085 
1 + 20.632 - 0.427 + 0.00041 
1 + 20.630 - 0.468 + 0.00126|.003 

+ 0.003 + 0.062-0.00122 
1 + 20.633 - 0.406 + 0.00004 

♦ We observe that in these two cases the estimated values of the roots are shown by 
the verification to be inaccurate. This should insure great care in making the verifi' 
cation. Th^ estimated values should nev^ b^ assumed to be accurate without verification 



202 ADVANCED ALGEBRA 

This indicates that a root is between .003 and .004. 

Verify, 1 + 20.630 - 0.468 + 0.00126|.004 

+ 0.004 + 0.083-0.00154 
1 + 20.634 - 0.385 - 0.00028 

This determines a root of (3) between .003 and .004. 

Thus one root of (1) correct to three decimal places is 1.213. 

The other root could be found similarly to be 1.229. 

EXERCISES 

Find all the real roots of the following equations correct to three decimal 
places. 

1. x^-1x-^1 = 0. 

2. 7x3 -8«2_i4x + 16 = 0. 

3. 2ic5_ 4x3-3x2 + 6 = 0. 

4. 4x4- 5x3 -8x + 10 = 0. 

5. 3x3 - 10x2 - 33x + 110 = 0. 



CHAPTEE XVIII 
DETERMINANTS 

185. Solution of two linear equations. We have already 
treated the solution of linear equations in two variables and 
stated (p. 47) the method of solving three or more linear equa- 
tions in three or more variables. This latter process is rather 
laborious and can be very much abridged and also developed 
more symmetrically by the considerations of the present chapter. 
Let us solve the equations 

a^x + b^ij = Ci, (1) 

a^x + b^y = Ca. (2) 

Multiply (1) by b^ and (2) by b^, and we obtain 

ai^2^ 4- bib^i/ = b2Ci ^. v 

Subtracting, we get (aib^ — a^bi) x = b^c^ — biC^, 

or if aA - ^2^1 ^ 0, ^^h^LIzh^. (3) 

ai/>2 — ^2^1 

Similarly, we obtain y = —^ ^' (4) 

We note that the denominators of the expressions for x and y 
are the same. This denominator we will denote symbolically by 
the following notation : ^, 

aibz — a^bi = , • _ 

^2 t*2 

The symbol in the right-hand member is called a determinant. 
Since there are two rows and two columns, this determinant is 
said to be of the second order. The left-hand member of the equa- 
tion is called the development of the determinant. The symbols 
Ui, bi, a2, ^2 ^^6 called elements of the determinant, while the ele- 
ments «! and &2 are said to comprise its principal diagonal. 

203 



(6). 



204 



ADVANCED ALGEBRA 



EuLE. The development of any determinant of the seconc 
order is obtained by subtracting from the product of the ele^ 
ments on the principal diagonal the product of the elements on 
the other diagonal. 



Thus 



Zi 



xyi - ziy ; 



12 3 
4 5 



10 - 12 = - 2. 



Evidently each term of the development contains one and only 
one element of each row and each column, that is, for instance, 
the letter a and the subscript 1 appear in each term of (5) once 
and only once. 

We can now rewrite the solution (3) and (4) of equations (1) 
and (2) in determinant form : 



X = 



Cl 


h 




ai Ci 


C2 


h 


; y = 


^2 ^2 


ai 


b. 


a^ bi 


a^ 


b. 




a^ b^ 



(6) 



It is noted that the numerator of the expression for x is formed 
from the denominator by replacing the column which contains 
the coefficients of x, a^ and a^,, by the constant terms c^ and c^. 
Similarly, in the numerator of the expression for y, b^ and b^ of 
the denominator are replaced by Cj and c^. 

One should keep in mind that a determinant is merely a sym- 
bolic form of expression for its development. In the case of 
determinants of the second order the introduction of the new 
notation is hardly necessary, as the development itself is simple ; 
just as we should scarcely need to introduce the exponential 
notation if we had to consider only the squares of numbers. It 
turns out, however, as we shall see, that we are able to denote by 
determinants with more than two rows and columns expressions 
with whose development it would be very laborious to deal. 

186. Solution of three linear equations. Let us solve the 
equations 



a^x + b^y 4- Ci« = c^i, 
a^x 4- b^2.y + ^2^ = ^2> 
<iz^ + b^y + Cg^ = c?8- 



(1) 

(2) 
(3) 



DETERMINANTS 



205 



Eliminating y from (1) and (2) and (1) and (3), we obtain 
(ai^g — a^h^x H- (b^c^ — hiCc^)z = dj)^ — d^b^j 
(a^bi — a^b^x + {c^b^ — b^c^z — d^b^ — d-J)^. 

Eliminating z, 

[iSHh — «2^i) (^3^1 — h(^i) — (ctsbi — a^bs) (b^Ci — ^iC2)]a; 

= (dibi — d^bi) (c^bi — bsCi) — (d^bi — dj)^) (b^Ci — b^c^} 

Developing, canceling, and dividing by bi, we obtain 

_ d^b^Cs + d^bsCi + ^3^1 ^2 — dib^c^ — dsb^c^ — d^b^c^ 

aib^c^ + a^b^c^ + a^b^c^ — a^b^c^ — a^b^c^ — a^Jy^c^ ^ 

Following the analogy of tbe last section, we write the denomi- 
nator 



ai^2^3 + «2^8^1 + «8^1^2 ~" %^3<'2 "~ «3^2Cl ~ «2^1^3 = 



«1 


l\ 


Cl 


^2 


K 


^2 


^3 


bs 


Cs 



(5) 



The right-hand member of this equation we call a determinant 
of the third order, and the left-hand member its development. 
As in the determinant of the second order, the elements a^, b^, Cg 
comprise the principal diagonal; each term of the development 
contains one and only one element of each row and each column, 
and all possible terms so constructed are included in the develop- 
ment. The signs of the terms of the determinant of the third 
order may be kept in mind by the following device. 
' Rewrite the first and second columns to the right of the deter- 
minant as follows : 



«! ^1 G^ 


<Zi &i 


% ^2 ^2 


^2 h 


«3 ^8 ^3 


% h 



The positive terms are found on the diagonals running down 
from left to right, the negative terms on the diagonals running 
up from left to right. 

The numerator of the fraction (4) expressed in determinant 
notation is 

«! bi Cj 

dz b^ Cg 
dn b^ Cg 



206 



ADVANCED ALGEBRA 



We can find similarly the values of y and z that satisfy equ£ 
tions (1), (2), and (3). The solutions of the equations in determi- 
nant form are as follows : 



d. 


^1 


^2 


d. 


b. 


d. 


h 


^3 


a^ 


K 


^1 


^2 


h 


<^2 


^3 


h 


Oz 



y = 



^1 


d. 


Ci 


^2 


d. 


C2 


^3 


d. 


Cz 


«1 


h 


^1 


^2 


h 


^2 


^3 


h 


Cz 



«i 


*i 


d. 


aj 


*s 


d. 


as 


Js 


d> 


a, 


h 


h 


a. 


h 


<'2 


as 


h 


Cs 



(6) 



The same principle observed on p. 204 for forming the deter- 
minants in the numerators of the expressions for x and y may be 
followed here. The determinant in the numerator of the expres- 
sion for ic, 2/, or z is found from the denominator by replacing the 
column that contains the coefficients of the variable in question 
by a column consisting of constant terms. Thus in the numerator 
of z we find the column d-^, d^^ d^ replacing the column Cj, Cg, Cg of 
the denominator. 



EXERCISES 



1. Find the value of the following determinants. 





3 2 1 






(a) 


4 6 2 
1 1 


• 




3 2 1 


Solution : 


4 6 2 








1 1 





4 


3 


1 


(b) 


1 


2 







6 


1 


1 




3 


4 


1 


(d) 





2 


4 




1 





6 





a X y 




(f) 


b c 
c b 


• 




c b 


(h) 


-c a 




-b -a 


o| 



= 18 + 4 + 0-6-8-0 = 8. 





2 





1 


(c) 


2 


1 







1 


2 







2 


1 


1 


(e) 


6 


3 


3 




4 


2 


3 





a 


b 


c 


(g) 


b 


c 


a 




c 


a 


b 







c 


b 


(i) 


c 





a 




b 


a 






DETERMINANTS 



207 



2. Solve the following equations by determinants. 



(a) 



2x + 32/ = 4, 



x-2y = l. 
Solution : Using the expressions (6), p. 206, we obtain 

-8 -3 _ n 
-4-3~y 



4 


3 


1 


-2 


2 


3 


1 


-2 



y = 



2 


4 


1 


1 


2 


3 


1 


-2 



(e) 



7 y = 12, 



7 a: + 2/ = 11. 



2x+5y=l, 2x + 72/:-l, a: + 4y = 2, 

^ ' 7a;+6?/ = 2. ^ -* Sx-9y = 2. ^ ' 2a;-3?/ = 

3. Solve the following equations by determinants. 

a: + y + 2 = 2, 
(a) X + 32/- 4 = 0, 

y - 2 2; = 6. 

Solution : Rearranging so that terms in the same variable are in a column, 
and supplying the zero coefficients, we get 

x+ 2/4- 2 = 2, 

a: + 32/ + 02 = 4, 

Ox+ y-2z = 6. 



By (6), p. 206, x 



2 


1 


1| 


4 


3 





6 


1 


-2 


J 


1 


1 


1 


3 








1 


-2 



-12+0 + 4-18 + 8 + 



18 



1 I /-^6 + + l 



0-0.+ 2 



= 6, 



1 


2 


1 




1 


4 










6 


-2 


-8+0+6-0+4+0 


1 


1 


1 


-3 


1 


3 










1 


-2 








1 2 
3 4 
1 6 



1 1 1 
13 
1-2 

Check: 6 + (-|) + (--L0) = 



,18 + + 2-0-6-4 
-3 



6-4 = 2. 



10 



10 



208 



\ AD 




2ic + 3y = 12, 
(b) 3a; + 2:2 = 1 

Sy + 4:Z 
x + y -z 
(d) x + z~ 
V + z- 

x + y + 
(f) 3a:-2z = 
5 y — 4 z = 0. 

X + 2/ + 2 = 9, 
(h) x + 2y + 3z = 14, 

jc + 3y + 62; = 20. 

.2x + .3y + .4z = 29, 
(j) .3x + .4y + . 521=38, 

.4a; + .5y + .6z = 51. 



ED AMe^EBRA 

7 ^lx-iy = 0, 
Ac) ix-iz = l, 
2; - i y = 2. 
a + 2y = 7, 
) 7x+92; = 29, 
y+8z = 17. 
X + y + 2 2; = 34, 
(g) X + 2 y + 2: = 33, 
2 X + y + 2: = 32. 
ax + 6y — C2; = 2 a6, 
(i) 6y + C2; — ax = 2 6c, 
C2; + ax — 6y = 2 ac. 
3x + 2y + 32: = 110, 
(k) 5 X + y - 4 2; = 0, 
2x-3y + 2; = 0. 



187. Inversion. In order to find the development of determi- 
nants with more than three rows and columns, the idea of an inver- 
sion is necessary. If in a series of positive integers a greater 
integer precedes a less, there is said to be an inversion. Thus in 
the series 12 3 4 there is no inversion, but in the series 12 4 3 
there is one inversion, since 4 precedes 3. In 1 4 2 3 there are two 
inversions, as 4 precedes both 2 and 3 ; while in 1 4 3 2 there are 
three inversions, since 4 precedes 2 and 3, and also 3 precedes 2. 

188. Development of the determinant. In the development of 
the determinant of order three we have 

%^2^3 + «2^3Cl + ^S^l^a — 0^8^2Cl — ^2^1^8 " «1^3^2- (1) 

the order of lettejps in each term the same as their 
in the principal diagonal (as we have done in the development 
above), it is observed that the subscripts in the various terms 
take on all possible permutations of the three digits 1, 2, and 3. 
The permutations that occur in the positive terms are 12 3, 
2 3 1, 3 1 2, in which occur respectively 0, 2, and 2 inversions. 
The permutations that occur in the subscripts of the negative 
terms are 3 2 1, 2 1 3, 1 3 2, in which occur respectively 3, 1, 
and 1 inversions. 




DETERMINANTS 209 

Thus in the subscripts of the positive terms an even number 
of inversions occur, while in the subscripts of the negative terms 
an odd number of inversions occur. This means of determining 
the sign of a term of the development we shall assume in general. 

When we have a determinant with 71 rows and columns it is 
called a determinant of the nth order. The development of such a 
determinant is defined by the following 

KuLE. The development of a determinant of the nth order is 
equal to the algebraic sum of the terms consisting of letters fol- 
lowing each other in the same order in which they are found in 
the principal diagonal hut in which the subscripts take on all 
possible permutations. A term has the positive or the negative 
sign according as there is an even or an odd number of inver- 
sions in the subscripts. 

This means of finding the development of a determinant is use- 
ful in practice only when the elements of the determinant are 
letters with subscripts such as in (2) below. When the elements 
are numbers we shall find the value of the determinant by a more 
convenient method. 

In this statement it is assumed that the number of inversions 
in the subscripts of the principal diagonal is zero. If this number 
of inversions is not zero, the sign of any term is + or — accord- 
ing as the number of inversions in its subscripts differs from the 
number in the subscripts of the principal diagonal by an even or 
an odd number. 

Since each term contains every letter a, b, ••-jk and also every 
index 1, 2, • • • , ri, one element of each row and column occurs in 
each term. 

In the determinant of the fourth order 

tti 61 Ci di 

a^ 62 C2 dz 

03 &3 C3 ^i 

a^ &4 C4 d^ 

the terms a^h^Cidi and a^h^c^di, for instance, have the minus sign, as 2 4 1 3 haa 
three inversions and 4 2 3 1 has five inversions; while the terms aih^c^d^ and 
a^b^Cidi have two and six inversions respectively and hence have the positive sign. 



210 



ADVANCED ALGEBRA 



189. Number of terms. We apply the theorem of permutar 
tions to prove the following 

Theorem. A determinant of the nth order has n ! terms in 
its development 

Since the number of terms is the same as the number of per- 
mutations of the n indices taken all at a time, the theorem follows 
immediately from the corollary on p. 145. 

190. Development by minors. In the development of the 
determinant of order three, p. 208, we may combine the terms as 
follows : 






<^l (pi^^Z — h<^2) — »2 (P\<^Z — h(^l) + 0^3 (^1^2 — h<^i) 



3 Cs 



+ Cli 



(1) 



IS 



We observe that the coefficient of a^ is the determinant that 
we obtain by erasing the row and column in which aj lies. A 
similar fact holds for the coefficients of a^ and a^. The determi- 
nant obtained by erasing the row and column in which a given 

element lies is called the minor of that element. Thus / 

the minor of cig- ^^ notice that in the above development by 
minors (1) the sign of a given term is + or — according as the 
sum of the number of the row and the number of the column 
of the element in that term is even or odd. Thus in the first 
term a^ is in the first row and the first column, and since 1 -f- 1 = 2, 
the statement just made is verified for that case. Similarly, aj is 
in the first column and the second row, and since 1 + 2 = 3 is 
odd, the sign is minus and the law holds here. The last term 
is positive, which we should expect since a^ is in the first column 
and the third row, and 1 + 3 = 4. The proof for the general 
validity of this law of signs is found on p. 215. 

The elements of any other row or column than the first may 
be taken and the development given in terms of the minors with 



DETERMINANTS 



211 





— (^2 




Cl 


+ ^'2 


«1 


Cl 


— ^2 


a^ 








^3 




dz 


<^Z 




as 



respect to such, elements. For instance, take the development 
with respect to the elements of the second row, 

d^ ^2 ^2 

The rule of signs is the same as given above ; that is, for 
instance, the last term is negative, as c^ is in the third column 
and the second row, and 2 + 3 = 5. By generalizing these con- 
siderations we may find the development of a determinant by 
minors by the following 

EuLE. Write in succession the elements of any row or column, 
each multiplied by its minor. 

Give each term a + or a — sign according as the sum of the 
number of the row and the number of the column of the element 
in that term is even or odd. 

Develop the determinant in each term by a similar process 
until the value of the development can be determined directly by 
multiplication. 

That this rule for development gives the same result as the definition given in 
§ 188 we have seen for a determinant of order three. The fact holds in general, as 
we shall prove (p. 215) . 

EXERCISES 

1. In the determinant of order four on p. 209 what sign should be pre- 
fixed to the following terms ? 

(a) Ct^hzO^di. 

Solution: c^^hza-idi = azhcidi. In 2 3 4 1 there are three inversions. The 
sign should be minus. 

(b) a^hcsdi. (c) aibic^ds. (d) b4Cidsa2- 
(e) ^8610402. (f) d2aic^b3. (g) 0208^164- 

. 2. Develop by minors the following and find the value of the determinant 
2 4 



(a) 



Solution : 



1 4 

1 6 

3 

2 

3 



= 3 
= 3.(6 



4) 



|1 
2(12 



4) + 3 (8 - 4) = 6 - 16 + 12 = 2. 



212 



ADVANCED ALGEBRA 






1 4 6 




(p) 


7 8 2 
1 3 1 
a b 




(e) 


d c 
e f 


• 





2 


1 





(c) 


2 





1 







3 


4 







a 


b 


(f) 


a 





b 




a 


6 






(d) 


5 7 

3 7 

-2 3 


2 

1 
- 1 


(g) 


a 6 
c 
d e 


• 



(h) 



Solution : Develop with respect to the elements of the first column, 



2 3 

3 \l 

4 2 
3 1 



112 
2 12 
12 3 

21 

2 31 

2 

3 






3 1 1 
2 12 
12 3 

1 2 



+ 4 



3 11 
112 
12 3 



2 3 

1 1 

2 3 



+ 1 



+ 1 



: s!)-('i; 



-2 



1 

2 3 

1 1 

1 2 



1 1 

1 2 

1 2 

1 



+ 1 



+ 2 



1 2 



(i) 



2 6 3 

i. ' 



= 2(-l + 2 + 0)-3(-3-2 + l) + 4(-3-l + l)-3(0-l+2) 
= 2 + 12 - 12 - 3 = - 1. 

1 2 



8 
6 
2 



9 
1 3 
1 8 
1 4 



ax 



1 1 
3 
3 



3 4 

1 1 

1 1 

3 3 



Hint. It is always advisable to develop with respect to the row or column that 
has a maximum number of elements equal to zero. 



(k) 



(m) 



(0) 



(q) 



a 








b 


b 


a 











b 


a 











b 


a 


2 


3 


4 


1 


2 


3 


3 


6 


2 


3 


1 


1 


2 


3 


2 


1 


X 


a 


b 




b 


X 


a 


. 


a 


b 


X 




X 











y 


X 











y 


X 











y 


X 











y 



(1) 



(n) 



(P) 



(r) 



«! 6i Ci di 

62 C2 di 



ai 
a2 62 
as 



C3 f^a 

^4 







63 
a^ 64 C4 
a gr 



e 


a 


/ 








a 


d 


c 


b 


X 


a 


b 


c 


X 


a 


b 


c 


X 


0, 


6 


c 



cs 




a 
c 
b 
a 

X 



DETERMINANTS 



213 



191. Multiplication by a constant. In this and the following 
sections we shall give a number of theorems on determinants 
which greatly facilitate their evaluation and which make a proof 
for the solution in terms of determinants of any number of linear 
equations in the same number of variables a simple matter. 

Theorem. If every element of a row or a column is multi- 
plied hy a number m, the determinant is multiplied by m. 

Suppose that every element of the first row of a determinant 
is multiplied by m. Since each term of the development contains 
one and only one element from the first row, every term is multi- 
plied by m, that is, the determinant is multiplied by m. 

Illustration. 

mbi bi ^3 
mci Cg C3 
= maiJg^a + fnazb^Ci -}- ma^biC^ — maib^c^, — ma^biC^ — ma^b^Cx 





«! a 


1 a. 






= m 


bx b^ ba 

Ci C2 C3 


• 




6 4 1 




Similarly, 


8 3 2 


= 






10 


4 1 





23 
24 
2-5 



= 2 



192. Interchange of rows and columns. We now prove the 

Theorem. The value of a determinant is not changed if the 
columns and rows are interchanged. 

Take for instance the determinant of order four. 



«! 


bi 


Cl 


di 




ai 


Oa 


as 


a^ 


aa 


b. 


C2 


d. 




bi 


b. 


bs 


b. 


ttg 


bs 


Cz 


d. 




Ol 


^2 


C3 


Ci 


a^ 


b. 


C4 


d. 




d. 


d. 


d. 


d. 



In each of these determinants the principal diagonal is the same, 
and hence the developments derived according to the statement on 



214 



ADVAl^CED ALGEBRA 



p. 209 will be the same for each determinant, since the terms wil 
be identical except for their order. The same reasoning is valic 
for any determinant. 

193. Interchange of rows or columns. We now prove the 

Theorem. If two columns or two rows are interchanged, the 
sign of the determinant is changed. 

Again let us take for example the determinant of order four and 
fix our attention on the first and second rows. We must prove that 



«! 


h 


Ci 


d. 




a^ 


h. 


^2 


d. 


as 


h 


Cl 


d. 


- 


«! 


W 


Cl 


d. 


as 


h 


Cs 


ds 




as 


h 


Gz 


ds 


a^ 


h 


Ci 


d. 




a^ 


b. 


C4 


d. 



In the first determinant the principal diagonal is a^bc^c^d^, while 
in the second the principal diagonal, a^b^^c^d^, is obtained from the 
principal diagonal of the first determinant by one inversion of 
subscripts. Hence this term is found among the negative terms 
of the first determinant. 

Since the only difference between the second determinant and 
the first is the interchange of the subscripts 1 and 2, evidently 
any term of the second is obtained from some term of the first 
by a single inversion. Thus if a single inversion is carried out in 
every term of the first determinant, we obtain the various terms 
of the second. But since this process changes the sign of each 
term of the first determinant (p. 213), we see that the second 
determinant equals the negative of the first. Similar reasoning 
may be applied to the interchange of any two consecutive rows 
or columns of any determinant. 

Consider now the effect of interchanging any two rows which 
are separated we will say by k intermediate rows. To bring the 
lower of the two rows in question to a position next below the 
upper one by successive interchanges of adjacent rows, we must 
make k such interchanges. To bring similarly the upper of the two 
rows to the position previously occupied by the other requires k-\-l 
further interchanges of adjacent rows. Hence the interchange of 
the two rows is equivalent to 2 A; + 1 interchanges of adjacent 



DETERMINANTS 215 

rows, the effect of which is to change the sign of the determinant, 
since 2 ^ + 1 is always an odd number. 

194. Identical rows or columns. This leads to the important 

Theokem. If a deteimiinant has two rows or two columns 
identical^ its value is zero. 

Suppose that the first and the second row of a determinant are 
identical. Suppose that the value of the determinant is the num- 
ber D. By § 193, if we interchange the first and second rows the 
value of the resulting determinant is — D. But since an inter- 
change of two identical rows does not change the determinant at 
all, we have j) ___ t) 

or 2 Z> = 0, that is, D = 0. 

Corollary. If any row (or column) is m times any other 
row (or column), the value of the determinant is zero. 

By § 191, the determinant may be considered as the product of 
m and a determinant which has two rows (or columns) identical. 
Hence this product equals zero. 

195. Proof for development by minors. On referring to the 
rule on p. 211 we observe that in order to show that the develop- 
ment by minors is the same as the development obtained by the 
definition on p. 209 we must prove the two following statements. 

First. The coefficient of any element in the development of a 
determinant (apart from sign) is the minor of that element. 

Second. Each element times its minor should have a -\- or 
a — sign according as the sum of the number of the row and the 
column of the element is even or odd. 

Consider the element ^j. 

First. Each term that contains ai must contain every other 
letter than a, and the indices of these letters must take on all 
permutations of the numbers 2, 3, ■ • -, n. This coefficient of ai 
contains then by definition (p. 210) all the terms of its minor. 

Second. Since in each term ai is in the first place, the only inver- 
sions in the subscripts are those among the numbers 2, 3, • • • , ?i. 



216 



ADVANCED ALGEBRA 



Hence the sign of each term in the coefficient of ai is positive or 
negative according as there is an even or odd number of inversions 
in its subscripts. Hence our theorem is established for the ele- 
ment %. 

Consider now any element, as d^ , which occurs in the fifth row 
and the fourth column. Interchange adjacent rows and columns 
until c?5 is brought into the leading position in the principal 
diagonal. This requires in all seven interchanges, three to get 
the c?6 in the first column, and then four to get it into the first 
row. This changes the sign of the determinant seven times, 
leaving it the negative of its original value. By the reasoning 
just given in the case of % the coefficient of d^ (which is now in 
the position previously occupied by ai) in the original determi- 
nant would be the minor of d^, except that the signs would 
all be changed. Hence the term consisting of d^ times its minor 
has the — sign, and the theorem is proved for this case. 

In general, to bring a term in the ith. row and kth. column to 
the leading position requires i — l-{-k — l = i-\-k — 2 inter- 
changes of adjacent rows or columns. This involves i -\- k — 2, 
or what amounts to the same thing, i + k changes of sign. Hence 
a positive or a negative sign should be given to an element times 
its minor according as i + ^ is even or odd. 

196. Sum of determinants. We now prove the fact that under 
certain conditions the sum of two determinants may be written in 
determinant form. The fact that the product of two determinants 
is always a determinant is extremely important for certain more 
advanced topics in mathematics, but the proof lies beyond the 
scope of this chapter. 

Theorem. If each of the elements of any row or any column con- 
sists of the sum of two numbers, the determinant may he written 
as the sum of two determinants. 

For example, we have to prove that 



«! + a'l 


^1 


Cl 




«! 


h 


Cl 




a\ 


^1 


Cl 


a^ + a\ 


h 


C2 


= 


^2 


h 


^2 


+ 


a'. 


h. 


Ci 


«» -f a's 


h 


^3 




^8 


h 


Cz 




«'s 


K 


Cs 



DETERMINANTS 



217 



Develop the first determinant by minors with, respect to the 
first column, where we symbolize the minors of Ui + a'j, a^ + a'2, 
ftg + a's by Ai, A^^ A^ respectively. 

We have 



by § 190, 



«! + a'l bi Ci 

^2 + a'2 ^2 ^2 

as + a's ^8 ^8 



«! 


^1 


Cl 




a\ 


h 


Cl 


a^ 


*2 


C2 


+ 


a\ 


h 


^2 


as 


^8 


Cs 




a\ 


h 


Cz 



= (»! -f a'O^l — (^2 + ^'2)^2 +(«8 + a'8)^8, 

by the distributive law, 

= aj^i — a^Az 4- 0.3^3 + a\Ay — a'2^2 + «'8^3> 



by § 190, 



The method of proof given is applicable to the case of any row 
or column of a determinant of order n. 

197. Vanishing of a determinant. For the solution of systems 
of linear equations we shall make use of the 

Theorem. If in the development of a determinant in terms 
of the minors with respect to a certain column (or row) the ele- 
ments of that column (or row) are replaced hy the elements of 
another column (or row), the resulting development equals zero. 

For example, we have 



«! 


h 


Ci 


d. 


a? 


h 


Ci 


d. 


as 


h 


Cz 


d. 


a^ 


h 


C4 


d. 



a-^A-i — a^A^ -p ^sA^ — a^A^j 



(1) 



where an A represents the minor of the a with the same sub- 
script. We have to prove that if we replace the a's, for example, 
by the b'Sj the result equals zero, that is, that 



biAi — b^A^ + ^3^3 — b^A^ — 0. 



(2) 



218 



ADVANCED ALGEBRA 



This expression (2) when written in determinant form evidently 
would have the same form as the left-hand member of (1) excej 
ing that the first column would consist of ^i , ^2 ? ^3 j ^4 • ^^ should 
then have two identical columns of the determinant, which would 
then equal zero (§ 194). Thus the development in (2) vanishes 
identically. This method of proof is perfectly general. 

Corollary. The value of the determinant is unchanged if the 
elements of any row (or column) are replaced hy the elements of 
that row (or column) increased hy a multiple of the elements of 
another row (or column). 

Thus, for instance. 



«! 


^1 


Ci 




^2 


h. 


^2 


= 


as 


h 


Cs 





«! 


-i-nb. 


h 


Ci 


^2 


+ nh^ 


h 


^2 


ds 


+ nh. 


h 


Cz 



The proof follows directly from §§ 191, 196, and the preceding 
theorem. 

198. Evaluation by factoring. If in a determinant whose 
elements are literal two rows or two columns become identical 
on replacing a by h, then a — b i^ a factor of the development. 
This appears immediately from § 160. 

Illustration. Evaluate by factoring 



(1) 



Since two columns become identical if a is replaced by h, a by c, 
or h by c, then we have as a factor of the development 

(a-b){b-c){c-a). (2) 

To determine whether the signs in this product are properly 
chosen, that is, whether the development should contain a — b oy 
b — a,we note that the term bc^ is positive in the development 
of (1) and also positive in the expansion of (2). Evidently there 
is no factor of (1) not included in (2). 



1 


1 


1 


a 


b 


c 


a^ 


b' 


c^ 



DETERMINAKTS 



219 



199. Practical directions for evaluating determinants. In 

j&nding the value of a numerical determinant the object is to 
reduce it to one in which as many as possible of the elements 
of some row or column are zero. One should ask one's self the 
following questions on attempting to evaluate a determinant: 

First. Is any row (or column) equal to any other row (or 
column) ? If so, apply § 191 for the case m = 0. 

Second. Are the elements of any row (or column) multiples of 
any other row (or column) ? If so, apply § 191. 

Third. If we add (or subtract) a multiple of the elements of 
one row (or column) to the elements of another, will an element 
be zero ? If so, apply § 197. 

EXERCISES 

Evaluate the following determinants. 



2 


3 


4 


3 


4 


5 


3 


2 


1 


2 


7 


6 





1 


8 


7 



Solution : We obseive that if we subtract each element of the first column 
from the corresponding element of the second column, the new second column 
has every element 1. A similar result is obtained by subtracting the last col- 
umn from the third column. Thus, by § 191, 



2 3 


4 


3 




2 


1 


1 


4 5 


3 


2 




4 


1 


1 


1 2 


7 


6 


"7 


1 


1 


1 


1 


8 


7 







1 


1 



= 0. 



4 3 12 

6 113 

4 2 12 

3 6 2 1 



Solution : Multiplying the last column by 2 and the whole determinant by \ 
does not change the value of the determinant (§ 191). Thus 



4 3 12 




4 3 14 


6 113 


_ 1 


6 116 


4 2 12 


~ 2 


4 2 14 


3 6 2 1 




3 6 2 2 



220 



ADVANCED ALGEBRA 



Subtracting the last column from the first column and developing, we 



obtam 



4 


3 


1 


4 




6 


1 


1 


6 


_ 1 


4 


2 


1 


4 


~2 


3 


6 


2 


2 





3 14 
116 
2 14 
16 2 2 



3 14 

1 1 6 

2 14 



Subtracting the last row from the first row, 



3 1 

1 1 

2 1 





1 6 
1 4 






-,-(-2) = l. 



a — d a 1 
b-d b 1 
c — d c 1 



a^ + b 
2ab 



2ab 


1 


a2 + 62 


1 


2a6 + 62 


1 



5. 



7. 



9. 



3 


3 


4 


2 


1 


1 


2 


1 


2 


2 


3 


1 


2 


1 


3 


2 






a; 


y 


2 


X 





y 


2 


y 


2 





« 


2 


y 


a; 






a 


1 








& 


1 


1 





c 


1 


2 





d 


1 


3 


3 



10. 



4 


6 


8 


3 


1 


1 


2 


1 


2 


3 


4 


1 


2 


1 


3 


4 



110 1 
10 11 
111 

1111 



1 


1 


1 





1 


1 





1 


1 





1 


1 





1 


1 


1 



11. 



a 


a 


a 


a 


a 


b 


b 


b 


a 


b 


c 


c 


a 


b 


c 


d 



12. 



2 


1 


1 


1 


1 


2 


1 


1 


1 


1 


2 


1 


1 


1 


1 


2 



13. 



3 


7 


16 


14 


6 


15 


33 


29 





1 


1 


1 


4 


2 


3 


1 



14. 



8 


2 


1 


4 


16 


29 


2 


14 


16 


19 


3 


17 


33 


39 


8 


38 



15. 



9 


13 


17 


4 


18 


28 


33 


8 


30 


40 


54 


13 


24 


37 


46 


11 



16. 



[2 


14 


16 


18 


2 


4 


6 


8 


4 


3 


2 


1 


3 


7 


11 


16 



DETERMINANTS 



221 



D, 



200. Solution of linear equations. Suppose that we have given 
n linear equations in n variables. We seek a solution of the equa- 
tions in terms of determinants. For simplicity, let n — 4:. Given 

a^x + hy + Ci« + d^w =/i, (1) 

a^x -f h^y + c^z + d^w =/2, (2) 

(^3^ + hy + CsZ + d^w =/8, (3) 

a^x + hy + C4^ + d^w =/4. (4) 

The coefficients of the variables taken in the order in which 
they are written^ift^ be taken as forming a determinant D, which 
we call the determinant of the system. Thus 
% hi Ci di 

(^2 t>2 ^2 ftg 
^8 ^3 ^3 ^8 

a^ b^ C4 d^ 

Symbolize by ^1, ^3, etc., the minors of %, 63, etc., in this deter- 
minant. Let us solve for x. Multiply (1), (2), (3), (4) by ^1, A^, 
Azj A^ respectively. We obtain 

A^a^x + A^biy + A^c^z + A^d^w = ^j/i, 
A^a^x + A^b^^y + ^3^2^ + A^d^w = ^2/2, 
^gaga: + A^b^y + ^3^3^ + ^af/gi^ = ^3/3, 
A^a^x -f- ^4^'4?/ 4- A^c^z + ^4C?4i^; = A^f^. 

If we add these equations, having changed the signs of the 
second and fourth, the coefficient of x is the determinant Z>, while 
the coefficients of y, z^ w are zero (§ 197). The right-hand mem- 
ber of the equation is the determinant D, excepting that the ele- 
ments of the first column are replaced by fufi, fs,f^ respectively. 
Hence 



X = 



/l 


h 


Cl 


d. 


A 


h 


Ca 


d. 


A 


h 


Cs 


d. 


A 


h 


^4 


d. 


«! 


h 


Cl 


d. 


^2 


h 


C2 


d. 


^8 


h 


Cz 


d. 


a^ 


h 


Ci 


d. 



222 



ADVANCED ALGEBRA 



In a similar manner we can show that the value of any variable 
which satisfies the equations is given by the following 

EuLE. The value of one of the variables in the solution of n 
linear equations in n variables consists of a fraction whose 
denominator is the determinant of the system and whose numer- 
ator is the same determinant, excepting that the column which con- 
tains the coefficients of the given variable is replaced by a column 
consisting of the constant terms. 

When D = 0, we cannot solve the equations unless the numer- 
ators in the expressions for the solution also vanish. 
Illustration. Solve for x 

ax -\-2by — 1, 

2by + ^cz = 2, 

S cz -{- 4: dw = Sj 

4 dw -{- 5ax = A. 

E-earranging, we obtain 

ax + 2 by = ly 

2by-^3ez =2, 

3 cz -{- 4: dw = 3j 

-{- 4:dW = 4:. 



D 



5 ax 

a 2b 

2b 3c 

3c 

5a 






v" 




Ad 


= 24 


4:d 





a 5 

5 c 

c ^ 

5a ^ ^ d 





1 


b 










1 


b 


24 


2 


b 


c 







1 


c 


3 





c 


d 




3 


c d 


X — 


4 








d 




4 


d 




a 


b 










a 


b 


24 





b 


c 










b c 








c 


d 







c d 




5a 








d 







-5b d 





1 


c 





-b 


3 


c 


d 




4 





d 




b 


c 





a 





c 


d 




-5b 





d 



DETERMINANTS 



223 





1 


c 





-h 


2 





d 




4 





d 




b 


G 





a 





C 


d 







5c 


d 



be 



ab 



G d 
5c d 



- 2 bed 
— 4 abed 



1_ 

2a 



201. Solution of homogeneous linear equations. The equa- 
tions considered in the previous section become homogeneous 
(p. 115) if /i =/2 =/3 =/, = 0. We have then 



a^x + b-^y -\- GiZ + diW = 0, 
azX -f bzy + G2% + dzW = 0, 
^3^5 + b^y + ^3^; + cZgW; = 0, 



(I) 



These equations have evidently the solution x = y = z = w =0. 
This we call the zero solution. We seek the condition that the 
coefficients must fulfill in order that other solutions also may 
exist. If we carry out the method of the previous section, we 
observe that the determinant equals zero in the numerator of 
every fraction which affords the value of one of the variables 
(§ 191). Thus if D is not equal to zero, the only solution 
of the above equations is the zero solution. This gives us 
the following 

Pkinciple. a system of n linear homogeneous equations in n 
variables has a solution distinct from the zero solution only 
when the determinant of the system vanishes. 

Whether a solution distinct from the zero solution always 
exists when the determinant of the system equals zero we shall 
not determine, as a complete discussion of the question would be 
beyond the scope of this chapter. 

Theorem. If x^, y^, z^, w^ is a solution of equations (I) 
and h is any number, then kx^, ky^, kz^, kw^ is also a solution. 



224 ADVANCED ALGEBRA 

The proof of this theorem is evident on substituting kx^ etc., 
in equations (I) and observing that the number A; is a factor of 
each equation. Thus if a system of n linear homogeneous equar 
tions has any solution distinct from the zero solution it has an 
infinite number of solutions. 

EXERCISES 
Solve for all the variables : 

x + y = a, a; + 3 y = 19, 

y + z = b, y-]-Bz = S, 

M + « = d, w + 3d = 11, 

V + x = e. V + 3ic = 15. 

z-\-y + w = a, Bx + y + z = 20, 

^z + w + x = b, -a; + 4y + 3w; = 30, 

*w + a; + y = c, '6jc + 2 + 3iy = 40, 

x-{-y-\-z = d. 8y + 3z + 6w; = 50. 

2/ + 2; + 5iy = ll, x — 2y + Sz — w = 6, 

^ x + z-\- 4:W = llj ^y-2z + Sw-x = 0^ 

'a: + y + 3M; = ll, ' z-2w-\-Sx-y = 0, 

x + z + Sy = 3S. w-2x + Sy - z = 6. 



7. 



x + y-\-z + w = 2i, x + y + z + w = 60, 

a; + 2y + 3z-9w = 0, g x + 2y + 3z -\- ^w = 100, 

3x-y -6z-\-w = 0, ' x-\-Sy -{■Qz-\-10w = 150, 

2x + 3y--4«- 610 = 0. x + 42/ + lOz + 20mj = 210. 



CHAPTEE XIX 
PARTIAL FRACTIONS 

202. Introduction. For various purposes it is convenient to 

f(x) 
express a rational algebraic expression ~j-\ ? § 11, as the sum of 

several fractions called partial fractions, which have the several 
factors of <^{x) as denominators and which have constants for 
numerators. If we write <j>(x) = {x — a)(x — P)--- (x — v), we 
seek a means of determining constants Aj B, ■, N such that for 
every value of x 

<l>{x) X — a X — p X — V ^ ^ 

If the degree of f(x) is equal to or greater than that of ^ (a;), 
we can write 

<i>(x) <}>(x) ^ ^ 

where Q (x) is the quotient and f^ (x) the remainder from dividing 
f(x) by <f>(x), and where the degree oi /^(x) is less than that of 
<^ (x). In what follows we shall assume that the degree of f(x) 
is less than that of <f> (x). In problems where this is not the case 
one should carry out the long division indicated by (2) and apply 
the principle developed in this chapter to the expression corre- 
sponding to the last term in (2). 

203. Development when </>(x) = has no multiple roots. Let 
us consider the particular case 

f(x) ^ x + 1 

<t> (x) (x -l)(x- 2) (x - 3)' 

We indicate the development required in form (1) of the last 
section, . . « ^ 



(x-l)(x-2)(x-3) x-1 x-2 x-S 
225 



226 ADVANCED ALGEBRA 

where A, B, and C represent constants which we are to deter- 
mine if possible. The question arises immediately, Are we at 
liberty to make this assumption? Are we not assuming the 
essence of what'we wish to prove, i.e. the form of the expansion? 
To this we may answer. We have written the expansion in form (1) 
tentatively. We have not proved it and are not certain of its 
validity. If, however, we are able to find numerical values of 
Ay B, and C which satisfy (1), we can then write down the actual 
development of the fraction in the form of an identity. 

If, on the other hand, we can show that no such numbers A,BjC 
satisfying (1) exist, then the development is not possible. 

Clear (1) of fractions, 

X + 1 = A(x - 2){x - ^)+ B{x -l){x - ^)-\- C (x -l)(x - 2) 
= {A + B-^C)x^-{6A +4.B-\-^C)x^QA+^B-\-2C. 

Since we seek values of A, B, and C for which (1) is identi- 
cally true for all values of x, equate coefficients of like powers of 
X in the last equation (Corollary II, p. 174). We obtain 

(2) 
(3) 
(4) 



(5) 
(2) 





A-\-B+ C = 

-5A _45-3C= 

6A +SB-\-2C = 


= 0, 
-.1. 


Add (4) 


to (3) and 


we obtain 




Adding we obtain 


A -B- C= 

A+B-hC = 

2A = 

A = 


= 2, 

:1 ^ 


From (^] 


1 and (4) we obtain 








B=-3j C = 


:2. 


Tlma 


x 


+ 1 


1 


XIlUo 


(x-l)(x 


-2)(a.-3) 


X-1 



4- 



As a check we might clear of fractions and simplify. If equar 
tions (2), (3), and (4) had been incompatible, we should have con- 
cluded that we could not develop the fraction in form (1). 



PARTIAL FRACTIONS 227 

We assume now for the general case 

^(x) = (x - a)(x - P)--(x - v), 

and that the n roots a, )3, • • •, v are all distinct from each other. 
Let us consider the expression 

f(x)_ A B N 

<^{x) X — a X — p X — V ^ 

where A, B, •", N are constants. Let us assume for the moment 
the possibility of expressing ~y^ in terms of these partial frac- 
tions. We shall now attempt to determine actual values A, B,--, 
iV which satisfy such an identity. K we multiply both sides of 
the identity by 

<t>(x) = (x-a)(x-P)-'(x-v), 
we obtain 

f(x) = A(x-P)'"(x-v) + B(x-a)--'(x-v) + -" 

^N(x-a)(x-P)"; 

f(x) is of degree not greater than n — 1, and consequently when 
written in the form of (1), p. 166, has not more than n terms. If 
we multiply out the righ1>hand member and collect powers of ar, 
we have an expression in a; of degree n — 1. By Corollary II, 
p. 174, this equation will be an identity if we can determine 
values ot Aj Bf "-,N which make the coefficients of x cm both 
sides of the equation equal to each other. Hence we equate 
coefficients of like jx>wers of x and obtain n equations linear in 
A, B, '",N which we can treat as variables. These equations have 
in general one and only one solution which we can easily deter- 
mine. The values of Aj B, • • • , iV obtained by solving these equa- 
tions we can substitute for the numerators of the partial fractions 
in (6). After making this substitution we can actually clear of • 
fractions the right-hand member of (6) and check our work by 
showing its identity with the left-hand member. 

There is no general criterion that we have applied to (6) to 
determine whether the n linear equations obtained lay equating 
coefficients of x have any solution or not. Hence in this general 



228 ADVANCED ALGEBRA 

discussion it should be distinctly understood that assumption (6) 
holds when and only when these equations are solvable. In 
any particular case we can find out immediately whether the 
equations are solvable by attempting to solve them. If the num- 
bers Aj By ' • ', N do not exist, the fact will appear by our inability 
to solve the linear equations. As a matter of fact, one and only 
one solution always exists under the assumption of this section. 

If in (6) we assume that several of the symbols A, B, ■'■, N stand for expres- 
sions linear in" x, as, for instance, ax + b, we should then have a larger number of 
variables to determine than there are equations. Under these circumstances there 
is an infinite number of solutions of the equations. Thus if we should seek to 

express =^-7— as the sum of partial fractions where the numerators are not con- 

stants but functions of z, we could get any number of such developments. 

We have the following 

EtTLE. Factor <\>{x) into linear factors, as 
(x — a)(x — P)"-{x — v). 
Write the expression 

</)(a?) X — a X — P x — v 

Multiply both sides of the expression by <i>{x)y equate coefficients 
of like powers of x, and solve the resulting linear equations for 
A,B,-^-,N. 

Replace A, B, •", N hy these values and check hy substituting 
for X some number distinct from a, ^, "•, v. 



EXERCISES 



Separate into partial fractions ; 



(x-l)(x-2)x 



/J.2 _ 2 A B f* 

Solution : Assume = 1 1 — . (1^ 

{x-l){x-2)x a;-lx-2a; ^' 

Multiply by (x - 1) (x - 2) x, 

x2 - 2 = ^(x - 2)x + J5(x - l)x + C(x - l)(x - 2), 

3fi-2 = {A+B-^C)x^-{2A+B + SC)z -f 2^. 



PARTIAL FKACTIOKS 229 



Equating coefficients of like powers of x, 
A + B + C = l, 
2A+B + SC = 0, 
2 O = - 2. 
Hence C = — 1. 

Solving 

we obtain 



Thus 



A + Bz 


= 2, 


2A + B = 


= 3, 


A = 


= 1 


J5 = 


= 1 


■ ' +- 


1 



x^-2 1.1 1 



(a;-l)(x-2)a; z-1 x-2 x 
Check : Let 

Substituting in (1), — - = — ~ -\ 



x = 


-A, 




-1 

-6 


r 

-■2 


*^.- 


1_ 
6~ 


1 
2 


-b' 



5+1=1. 

6 -6 



2. 


x-1 

x^ -\-Sx-^2 


A 


1 




3 a;2 _ 2 X - 8 


(i 


4a;2 




{x^-i){x-l) 


R 


a;2_3a; + l 



3. " + '' 



1. 



2a;2-5x-3 




5x 




6x2-5x-l 




2x2-1 




(x2 + 3a; + 2)(x- 


-1) 


a;2 4-4 





9. 

(X - 1) (X - 2) (X - 3) (X - 2) (X + 2) (X - 1) 

204. Development when ^ (a?) = O has imaginary roots. In 

the preceding section no mention has been made of any distinc- 
tion between real and imaginary values oi a, p, ■ -- , v. In fact 
the method given is valid whether they are real or imaginary. It 
is, however, desirable to obtain a development in which only real 
numbers appear. 

Let us assume the development 

iS=-^ + -^ + - + ^^ + -^' (1) 

^ {x) X — a X — p X — fji X — V 

where let us suppose that /x and v are the only pair of conjugate 
imaginary roots of <^ (.r), m and n being conjugate complex numbers. 



230 ADVANCED ALGEBRA 

Let fi = a-\-ib, v = a — ib. 

Then adding the corresponding terms of (1), we obtain 

m n _ X (m -{- n) — a (m + n) b(m — n) 

x — a — ib x — a + ib (x — a)^ -{- b^ (x — af-\-b'^ ^ ^ 

Since /a and y are the only imaginary roots of <f> (x) = 0, the last 
term of (2) is real, as is also the entire right-hand member (§ 152). 
Hence, letting the numerator 

x(7}i-\-n)— a(m-\-n)-\- ib (m — n) — Mx + N, 

we have the development 

/M^^4_ _B_ ■ Mx + N 

<^(x) x-a x- ft "^ (ic _ a)2 -|. J2 W 

By complete induction we can establish this form of numerator 
where there is any number of pairs of imaginary roots of <f>(x) = 0. 
^e have proved the form (3) where there is one pair of imagi- 
nary roots. Assuming the form where there are k pairs, we can 
prove it similarly where there are A: + 1 pairs. Hence we have the 

Theorem. If <l>(x) is facto7^ahle into distinct linear and quad- 
ratic factors, but the quadratic factors are not further reducible 

fix) 
into real * factors, then is separable into partial fractions 

of the form ^^ ^ 

A B Mx + N 

+ ;;+•••+ o ■ ■ > 



X — a X — P a? -\- ^x + V 

where x^ + fix -\- v is an irreducible quadratic factor of <t>(x). 

This theorem is of course true only under the condition that 
the linear equations obtained in the process of determining the 
constants are solvable. It turns out, however, that in this case 
as in § 203 the linear equations obtained always have one and 
only one solution provided that the roots are all distinct. 

* A real /actor is one whose coefficients are all real numbers. 



PARTIAL FRACTIONS 



231 



EXERCISES 



Separate into partial fractions : 



x2+ 1 



(X - 1) (x2 + X + 1) 
Solution : Assi^me 



X2+1 



A Bx + C 



...^ (X - 1) {X2 + X + 1) X - 1 X2 + X + 1 

Multiply by (x -^{af^ + a; + Ij, - ' 

x2 + 1 = ^ (x2 + x + 1) + (Bx + 0) (« - 1). 

Collecting like powers of x, 

3fi + l = {A + B)x^-{- {A- B+ C)x + A-C. 

Equating coefficients of x, 

A + B = l, 

A-B-{-C = 0, 

A-C = l. 

Add (2) and (3) and solve with (1), 

2A-B=1 

A-{-B=:l 



Substituting in (1), 
Substituting in (3), 



SA = 2, 

A = l 

B = h 



> 



(1) 
(2) 
(3) 



Thus 



X2 + 1 



x-1 



(x-l)(x2 + x + l) 3(x-l) 3x2 + 3x + 3 

Check: Let x = — 1. 

2 2 -2 



Substituting, 
Reducing, 

„ X2 + a + 1 



2-1 3 

6 3 



2 3-3+3 
1. 
x2 + l 



x3 + 4x 

x2 + 4 


x8-2x2 + 3x- 

X 


2 


(X + 3) (2 x2 - X 

X2 + 1 


-4) 



X* + X2 + 1 

5x^-1 

x4 + 6x2 + 8* 

1 



(X - l)(3x2 + x + 6) 



x3 + 3x2-2x-16 
Hint. Factor by synthetic divi- 
sion (see p. 178) . 



232 ADVANCED ALGEBRA 

205. Development when ^ (x) = (x — a)~. In this case the 
method given in the previous sections fails, as the equations for 
determining the values of the numerators are incompatible. If 

^®^®* f{x) ^ a,x-' + a,x-' +... + a^_^ 

<f>(x) (x-ay ' ^ ^ 

we can separate into partial fractions as follows. 

Let X — a = y, that is, x = y -\- a, and substitute in (1). We 
obtain after collecting powers of y 

_ = 1 — -^ f_ — _, 

y y y y 

where the ^'s are constants. Eeplacing yhyx — a,wQ have th^ 
following development : 

A, 



/(f) = ^0 I ^1 I ^'^ I ....). 

(x — a)" X — a (x — ay (x — ay (x — ay 



EXERCISES 

Separate into partial fractions : 

- x2 + 2 a; - 1 

(X-3)8 

Solution : Assume — -— = H — ■ -(- — -. (1) 

(X - 3)8 jc - 3 (X - 3)2 (X - 3)8 ^ ' 

Multiply by (x - 3)8, 

x2 4- 2 X - 1 = ^ (x2 - 6x + 9) + 5(x - 3) + C 

Collecting powers of x, 

= Ax'^ + {B-6A)x-\-9A-BB-\-C. 
Equating coefficients of x, 

^ = 1, 

5-6^ = 2, 

9A-SB + C = -1. 

Solving, B = 8, C = U. 

„ x2 + 2x-l 1,8 14 

Hence — ' = 1 \- 



(X - 3)8 X - 3 (X - 3)2 (X - 3)8 

Check : Let x = 1. 

^ ^ ,.,,.. ,,, 2 1 8 14 1 „ 14 

Substituting in (1), __ = — - + --_=_- + 2-— , 

1 1 



PARTIAL FRACTIONS 233 



2. ^^^.. 3. 



(x-2)3 (2a; + 1)2 (cc - 4)3 

g x^ + x + 1 g a;- g - 2 a;^ + 3 x + 1 

(2x-l)*" ' (ax + 6)2* * (3x-2)3 

206, General case. When <f>(x)= has real, complex, and 
multiple roots, we may use all the previous methods simultane- 
ously. Hence for this case we assume the expansion 

f(^ 

(x — a)--- (Xx^+ fxx -\- v) ■ " (x — tY 



X — a Ax2-f fjiX -\- V X — T (£c — t)' 



EXERCISES 

Separate into partial functions : 
- g^ + 2x2 + i8x-18 
' (x-l)(x2 + x + l)(x-2)2' 

Solution : 

x* + 2x2 + 18x-18 A Bx-VG D . E 



(X - 1) (x2 + X + 1) ^ - 2)2 X - 1 x2 + X + 1 X - 2 (X - 2)2 

= ^(x2 + X + 1) (x - 2)2 + {Bx + C) (X - 2)2 (X - 1) 
+ i)(x-l)(x-2)(x2 + x + l) 

+ JSJ(X-1)(X2 + X + 1) 

= {A-hB+D)x^-\-{-3A-6B + C-2D-\-E)x^ 
-{-{A + 8B-5C)x2 + (-4E + 8C'-D)x 
+ (4 ^ - 4 C + 2 D - ^). 

Equating coefficients of like powers of x, 

A+ B + D =1, (1) 

-SA-6B-\- C-2D + E = 0, (2) 

A-^8B-5C =2, (3) 

- 4 B + 8 C - D =18, (4) 

4 A -4C + 2D-^ = -18. (6) 

Adding (2) and (5), (1) and (4), we have, together with (3), 

^_5E-3C' = -18, (6) 

A-3B-\-SC = 19, (7) 

A-\-SB-6C = 2. (8) 



234 ADVANCED ALGEBRA 

Adding all three equations, we find 

3^ = 3, or A = l. 

Substituting in (3) and (7) and solving, we find C = S, B = 2. Substituting 
in (1), we find D = - 2. Similarly, from (6), ^ = 6. 

x* + 2a;2 + 18x-18 1 , 2a; + 3 2 , 6 
Thus — — - = + — 4- 



(x-l)(x2 + a; + l)(x-2)2 x-1 x^ + x + 1 x-2 (a; -2)2 

Check: Let x = — 1. 

go J 2 2 6 

Substituting, -^-^ = __ + ____ + _, 

¥ = 2i-i = -V-. 
2. -^!±^. 3. 

X{X- 1)3 

. X3 -f 2 X2 + 1 _ 

4. o. 

X(X-1)3 

g 2x3 + x + 3 y 

(X2 + 1)2 

8 ^~^ 9 

■ (x + l)2(x + 2)' 

10. ^^^ + ^ . 11. 

(a:2-l)(x + 2) 

12. ^^-^^ + ^ . 13. 

(x - 8) (X - 9) 

.. 2x2 -3x- 3 .- 

14. 15. 



16. "^^ — tl 17. 

(2x-3)(6x2-6x + l) (a;2_3a;4.2)(x-3) 



l-x* 




5X + 12 




x(x2 + 4) 




43X-11 




30(x2-l) 




X3 - X - 1 




X4-16 




x2 + 6x-8 




x3-4x 




2 




(x2 + X + 3) (2 a 


• + 1) 


x3 4-2x2-3x 


+ 1 


(x + 3)(x2-4x 


+ 5) 


13 -5x 





CHAPTEE XX 
LOGARITHMS 

207. Generalized powers. If h and c are integers, we can easily 
compute h''. When c is not an integer but a fraction we can com- 
pute the value of ¥ to any desired degree of accuracy. Thus if 
6 = 2, c = I, we have 2^ = V2* = VS, which we can find to any 
number of decimal places. If, however, the exponent is an irrar 
tional number as V2, we have shown no method of computing 
the expression. Since, however, V2 was seen (p. ^5) to be the 
limit approached by the sequence of numbers 

1, 1.4, 1.41, 1.414, ••, 

it turns out that 5^^ is the limit approached by the numbers 



The computation of such a number as S^"*^ would be somewhat 
laborious, but could be performed, since 5^-^^ = 5**^ = V5^*\ Thus 
it is a root of the equation x^^ = S^'*^ and could be found by 
Horner's method, p. 197. 

We see in this particular case that 5^^ is the limit approached 
by a sequence of numbers where the exponents are the successive 
approximations to V2 obtained by the process of extracting the 
square root. In a similar manner we could express the meaning 
of 5", where i is a positive integer and c is any irrational number. 

Assumption. We assume that the laws of operation which we 
have adopted for rational exponents hold when the exponents are 
irrational. 

Thus 6c.6d= 6c+d^ ^= 6e-d, {Tbd)c= {ho)d= bcd^ 

where c and d are any numbers, rational or irrational. 

236 



236 ADVANCED ALGEBRA 

208. Logarithms. We have just seen that when h and c are 
given a number a exists such that ¥ = a. We now consider the 
case where a and h are given and c remains to be found. Let 
a = 8, ^ = 2. Then if 2*^ = 8, we see immediately that c = 3 satis- 
fies this equation. If a = 16, 5 = 2, then 2" = 16 and c = 4 is the 
solution. If a = 10, 6 = 2, consider the equation 2" = 10. If we 
let c = 3, we see that 2^ = 8. If we let c equal the next larger 
integer, 4, we see 2* = 16. If then any number c exists such that 
2*^ = 10, it must evidently lie between 3 andj:. To prove the exist- 
ence of such a number is beyond the^^ope of this chapter, but 
we make the following 

Assumption. There always exists a real number x which 
satisfies the equation 

b^ = a, (1) 

where a and h are positive numbers, provided b ^ 1. 

Since any real number is expressible approximately in terms of 
a decimal fraction, this number x is so expressible. 

The power to which a given number called the base must be 
raised to equal a second number is called the logarithm of the 
second number. 

In (1) X is the logarithm of a for the base b. 

This is abbreviated into 

X = \ogf,a. (2) 

Thus since 

28 = 8, 102 = 100, 3- 2 = J, 40 = 1, 
we have 

3 = logaS, 2 = logiolOO, - 2 = logg j, = log4l. 

The number a in (1) and (2) is called the antilogarithm. 

EXERCISES 

1. In the following name the base, the logarithm, and the antilogarithm, 
and write in form (2). 

(a) 36 = 729. 

Solution : 3 = base, 6 = logarithm, 729 = antilogarithm, logs 729 = 6. 

(b) 2* = 16. (c) 38 = 27. 



LOGARITHMS 237 

2. Find the logarithms of the following numbers for the base 3 : 

81, 243, 1, i, ij. 

3. For base 2 find logarithms of 8, 128, |, ^V 

4. What must the base be when the following equations are true ? 
(a) log 49 = 2. (b) log81 = 4. 

(c) log 225 = 2 . (d) log 625 = 4. 

209. Operations on logarithms. By means of the law expressed 
in the Assumption, § 207, we arrive at principles that have made 
the use of logarithms the most helpful aid in computations that 
is known. 

Theorem I. The logarithm of the prodicct of two numbers is 
the sum of their logarithms. 

Let log^a. = a;, 

logfeC = y. 
Then by (1) and (2), p. 236, h^ = a, 

¥ = c. 
Multiply (by Assumption, § 207), 

or by (1) and (2), \og^a-c = x -\- y. 

Theorem IL The logarithm of the nth power of a number is 
n times the logarithm of the number. 



Let 


logi,a = Xy 


or 


b' = a. 


Raise both sides to the nth. power, 




(b^f. = b'^ = a% 


or 


logft a" = nx. 


Example. 


Iogiol00 = 2, 




logio 1000 = 3. 


By Theorem I, 


logio 100,000 = 5, 


which is evidently true, 


since 10^ = 100,000. 




238 ADVANCED ALGEBRA 

Theorem III. The logarithm of the quotient of two 
is the difference between the logarithms of the numbers. 

Let logj, a = Xj 

log6C = 2/. 
Then b'' = a, 

b^ = c. 

Dividing, ^,«-y = -, 

c ^^ 

, a 
or log;, -==x — y. 

G 

Theorem IV. The logarithm of the real nth root of a nur.t- 
ber is the logarithm of the number divided by n. 

Let logj, a = Xj 

or b"" = a. 

1 X 

Extract the nth root, (b'^y = ^^ = -Va, 

or logj -Va = - • 

n 

EXERCISES 

Given logio2 = .301, logio5 = .699, logio? = .8451, find 

1. log(-s/f^- V5).* 
Solution : 

By Theorem I, log ( v^T^ • VS) = log v^ + log V5. 
By Theorems III and IV, = f log 7 + ^ log 5. 

Now f log 7 = f (. 8451) = . 50706, 

and i log 5 = |_(.699) = .349^ 

Adding, | log 7 + | log 5 = log ( Vt^ . VB) = .85656 

2. log 40. 3. log 28. 
Hint. Let 40= 8-5= 28.6. 

4. log 140. 5. log V280. 

6. log V'35. 7. log ( V8 . v^ . V^). 

8. log ( V5 . 78). 9. log ("t/Ie . Vli . 4^700), 

• Where no base is written it is assumed that the base 10 is employed. 



LOGARITHMS 239 

210. Common system of logarithms. For purposes of compu- 
tation 10 is taken as a base, and unless some other base is indi- 
cated we shall assume that such is the case for the rest of this 
chapter. We may write as follows the equations which show the 
numbers of which integers are the logarithms. 

Since 10^ = 100,000 we have log 100,000 =A 



10* = 10,000 


log 10,000 


= 4. 


108 ^ IQQQ 


log 1000 


= 3. 


102 ^ IQQ 


log 100 


= 2. 


10^ = 10 


log 10 


= 1. 


10« = 1 


logl 


= 0. 


io-» = .i 


log.l 


= -1. 


10-2 = .01 


log .01 


= -2. 


10- « = .001 


log .001 


= -3. 


etc. 


etc, 





Assuming that as x becomes greater log x also becomes greater, 
we see that a number, for example, between 10 and 100 has a 
logarithm between 1 and 2. In fact the logarithm of any number 
not an exact power of 10 consists of a whole-number part and a 
decimal part. 

Thus since IQS < 3421 < lO^, 

log 3421 = 3. + a decimal. 
Since 10- 3 < .0023 < 10-2, 

log .0023 = - 3. + a decimal. 

The whole-number part of the logarithm of a number is called 
the characteristic of the logarithm. 

The decimal part of the logarithm of a number is called the 
mantissa of the logarithm. 

The characteristic of the logarithm of any number may be seen 
from the above table, from which the following rules are imme- 
diately deduced. 

'f The characteristic of the logarithm of a number greater than unity 
is one less than the number of digits to the left of its decimal point. 

Thus the characteristic of the logarithm of 471 is 2, since 471 is between 100 
and 1000; of 27.93 is 1, since this number is between 10 and 100; of 8964.2 is 3, 
since this number is between 1000 and 10,000. 



240 ADVANCED ALGEBRA 

The characteristic of the logarithm of a number less than 1 
is one greater negatively than the nwmher of zeros ^preceding the 
first significant figure. 

Thus the characteristic of the logarithm of .04 is — 2 ; of .006791 is — 3 ; of 
.4791 is - 1. 

It must constantly be kept in mind that the logarithm of a 
number less than 1 consists of a negative integer as a character- 
istic plus a positive mantissa. To avoid complication it is desir- 
able always to add 10 to and subtract 10 from a logarithm when 
the characteristic is negative. Thus, for instance, instead of writ- 
ing the logarithm - 3 + .4672 we write 10 - 3 + -4672 - 10, or 
7.4672 — 10. This is convenient when for example we wish to 
divide a logarithm by 2, as by Theorem IV, § 209, we shall wish 
to do when we extract a square root. Since in the logarithm 
— 3 -f .4672 the mantissa is positive, it would not be correct to 
divide — 3.4672 by 2, as we should confuse the positive and 
negative parts. This confusion is avoided if we use the form 
7.4672 - 10, and the result of division by 2 is 3.7336 - 5, or 
8.7336 — 10. The actual logarithm which is the result of this 
division is — 2 + .7336. 

Theorem. Numhers with the same significant figures which 
differ only in the position of their decimal points, have the same 
mantissa. 

Consider for example the numbers 24.31 and 2431. 

Let 10^ = 24.31. 

. Then a = log 24.31. 

If we multiply both numbers of this equation by 100, we have 
10^10^ = 10^+2^2431, 
or x-\-2 = log 2431. 

Thus the logarithm of one number differs from that of the other 
merely in the characteristic. In general numbers with the same sig- 
nificant figures are identical except for multiples of 10. Hence their 
logarithms differ only by integers, leaving their mantissas the same. 

Thus if log47120. = 4.6732, log47.12 = 1.6732, and log .004712 = - 3.6732, or 
7.6732 - 10. 



LOGARITHMS 241 



EXERCISES 



By §209, 


log V600 = 1.38905 




2. log .06. 


3. log (210)3. 


4. log 


5. log(4.2)4. 


6. log 

"^2.1 


7. log 


a , 567 


Q , 13.23 

9. log 

^ 1.28 





If log 2 = .3010, log 3 = .4771, log7 = .8451, find 
1. logVOOO. 

Solution : log V600 = log V20 • 30 = ^ log 20 + i log 30. 

By the preceding theorem, log 20 = 1.3010, log 30 = 1.4771. 
i log 20= .6505 
I log 30= .73855 



(70)3 
324 



211. Use of tables. A table of logarithms contains the man- 
tissas of the logarithms of all numbers of a certain number of 
significant figures. The table found later in this chapter gives 
immediately the mantissas for all numbers of three significant 
figures. In the next section a method is given for finding the 
mantissa for a number of four figures.* Hence the table is called 
a four-place table. Before every mantissa in the table a decimal 
point is assumed to stand, but in order to save space it is not 
written. To find the logarithm of a number of three or fewer 
significant figures we apply the following 

Rule. Determine the characteristic hy rules in § 210. 

Find in column N the first two significant figures of the num- 
ber. The mantissa required is in the row with these figures. 

Find at the top of the page the last figure of the number. 
The mantissa required is in the column with this figure. 

When the first significant figure is 1 we may find the loga- 
rithm of any number of four figures by this rule from the table 
on pp. 248, 249 if we find the first three instead of the first two 
figures in column N. 

Thus the log 516. = 2.7126, log .00281 = - 3.4487, log 7400. = 3.8692, 
log 600. = 2.7782, log 50. = 1.6990, log 4.00 = .6021. 



242 ADVANCED ALGEBRA 

EXERCISES 
Find the logarithms of the following : 

1. 3. 2. 303. 3. .024. 

4. 347. 6. .0333. 6. 1.011. 

7. .202. 8. .0029. 9. .0001. 

10. .00299. 11. 68400. 12. .0201. 

212. Interpolation. We find by the preceding rule that, 
log 2440 = 3.3874, while log 2450 = 3.3892. If we seek the loga- 
rithm of a number between 2440 and 2450, say that of 2445, 
it would clearly be between 3.3874 and 3.3892. Since 2445 is just 
halfway between 2440 and 2450, we assume that its logarithm is 
halfway between the two logarithms. To find log 2445, then, we 
look up log 2440 and log 2450, take half (or .5) their difference, 
and add this to the log 2440. This gives 

log 2445 = 3.3874 + .5 x .0018 = 3.3883. 
If we had to find log 2442 we should take not half the difference but 
two tenths of the difference between the logarithms of 2440 and 
2450, since 2442 is not halfway between them but two tenths of the 
way. This method is perfectly general, and we may always find the 
logarithm of a number of more than three figures by the following 

Rule. Annex to the proper characteristic the mantissa of 
the first three significant figures. 

Multiply the difference between this mantissa and the next 
larger mantissa in the table (called the tabular difference and 
denoted by D) by the remaining figures of the number preceded 
by a decimal point. 

Add this product to the extreme right of the logarithm of the 

first three figures, rejecting all decimal places beyond the fourth. 

In this process of interpolation we have assumed and used the principle that 
the increase of the logarithm is proportional to the increase of the number. This 
principle is not strictly true, though for numbers whose first significant figure 
is greater than 1 the error is so small as not to appear in the fourth place of the 
logarithm. For numbers whose first significant figure is less than 2 this error 
would often appear if we found the fourth place by interpolation. For this reason 
the table on pp. 248, 249 gives the logarithms of all such numbers exact to four 
figures, and in this part of the table we do not need to interpolate at all 



LOGARITHMS 243 



EXERCISES 

Find the logarithms of the following : 
1. 63.924. 

Solution: log 63.9 = 1.8055 

2 



Tabular difference = 



4. 62230. 




7. 20060. 




10. 9.999. 




13. 5.7828. 




16. 3.1416. 




log 275 =2.4393 


D= 16 


6 
2.4399 


.4 
6.4 



log63.924 = 1.8057 iL 

1.68 
We add 2 to 1.8055 rather than 1, because 1.68 is nearer 2 than 1. In general 
we take the nearest integer. 

2. 269.4. 3. 1001. 

5. 392.8. 6. 9.365. 

8. .4283. 9. .3101. 

11. 82.93. 12. .05273. 

14. .003011. 15. .002156. 

17. 276.4 X 1.463. 
Solution : log 275.4 = 2.4399 

log 1.463 = 0^652 
By Theorem II, § 209, log (275.4 x 1.463) = 2.6051 

18. 374.3 X 1396. 19. 1.46 x 237.2. 
20. 469.1 X 63.92. 21. 47320. x .8994. 

22. :5?!?1. 

^8-^^ log .0372 =8.5705 -10 /)= 12 

Solution: log. 03724 = 8.5710 - 10 5 _A 

log38.46 =L585^_ ^.^^J^'^ ,M 

6.9860 - 10 7 .6 

1.5850 7^ 

23 5:^. 24 1^51??. 

.2364 ' 5.128 

213. Antilogarithms. We can now find the product or quotient 
of two numbers if we are able to find the number that corresponds 
to a given logarithm. 

For this process we have the following 

KuLE. If the mantissa is found exactly in the table, the first 
two figures of the corresponding number are found in the column 
N of the same row, while the third figure of the number is found 
at the top of the column in which the mantissa is found. 

Place the decimal point so that the rules in § 210 are fulfilled. 



244 ADVANCED ALGEBRA 

EXERCISES 

Find the antilogarithms of the following: 

1. 3.7419. 
Solution: We find the mantissa .7419 in the row which has 55 in coT 

umn N. The column in which .7419 is found has 2 at the top. Thus the 
significant figures of the antilogarithm are 552. Since the characteristic is 3, 
we must by the rule in § 210 have four figures to the left of the point. 
Thus the number sought is 5520. 

2. 1.3874. 3. 2.7050. 4. .6785. 
5. 2.8414.* 6. 5.8831. 7. 1.5752. 
8. 9.9112 - 10. ■ 9. 3.7251. 10. 6.3997. 

If the mantissa of the given logarithm is between two man- 
tissas in the table, we may find the antilogarithm by the following 

EuLE. Write the number of three figures corresponding to the 
lesser of the two mantissas between which is the given mantissa. 

Subtract this mantissa from the given mantissa, and divide 
this number by the tabular difference to one decimal place. 

Annex this figure to the three already found, and place the 
decimal point as the rules in § 210 require. 

It should be kept in mind that we may always add and subtract 
any integer to a logarithm. This is useful in two cases : 

First. When we wish to subtract a larger logarithm from a 
smaller ; 

Second. When we wish to divide a logarithm by an integer 
that is not exactly contained in the characteristic. 

Both these processes are illustrated in exercise 2 (1) following. 

EXERCISES 

1. Find the antilogarithms of the following : 

(a) 2.3469. 

Solution : The mantissa 3469 is between 3464 and 3483. Hence D = 19. 

The mantissa 3464 corresponds to 222. To find the fourth significant figure 
of the antilogarithm, divide 3469 - 3464 = 5 by D = 19. Since 5 -^ 19 = .26, 
we annex 3 to 222. Hence the antilogarithm = 222.3. 

* We write - 2 + .8414 in the form 2.8414 to save space and at the same time to recall 
the fact that the mantissa is positive. 



LOGARITHMS 



245 



(b) 4.3147. (c) 1.5271. (d) 1.4216. 

(e) 1.6423. (f) 2.8791. (g) .7214. 

2. Perform the following operations by logarithms. 
1375 X .06423 



(a) 



76420 



Solution : 


log 1375= 3.1383 
log. 06423= 8.8077- 10 




Adding (Theorem I, 


§ 209), 11.9460 - 10 
log 76420= 4.8832 




Subtracting (Theorem III, § 209), log result = 7.0628 - 10 






result = .001156. 




(b) (11)8. 


(c) iUUY- (d) 5871 -, 


- 9308. 


(e) (H)«. 


(f) (3f I)*-". (g) 7066 -. 


-5401. 


(h) 8308 X .0003769. 


(i) 3410 X .008763. 




8.371 X 834.6 


,, , 37.42 X 11.21 




^■"^ 7309 


^^^ BBAl ■ 




^1) 8/87xV7194 
\ 98080000 








Solution : 


log 87= 1.9395 




By Theorem IV, § 209, i log 7194 = 1.9285 




Adding, 


= 13.8680 - 10 
log 98080000= 7.9916 




By Theorem III, § 209, 3)25.8764 - 30 






log result = 8.6255 - 10 






result = .04222. 





Since in the subtraction in this problem we have to subtract 7 from 3, 
we add and subtract 10 to the minuend to avoid a negative logarithm. Since 
in the division by 3 we would have a remainder in dividing — 10 by 3, we 
add and subtract 20 so that 3 may be exactly contained in 30, the negative 
part of the logarithm. 



(m) ^. 


(n) V:06. 


(0) ^(.043)8. 


(P)^^. 


(q) (.21)§. 


(r) m V'lOO. 


(s) •^:o3. 


(t) ^100. 


(u) V(l. 563)3. 


(v) V.00614. 


(w) VHi- 


(X) ^0.9 VI!- 



(y) ■>J^-v^ 



(z) >y.47 VM- 



214. Cologarithms. The logarithm of the reciprocal of a num- 
ber is called its cologarithm. When a computation is to be made 



246 



ADVANCED ALGEBRA 



in which several numbers occur in the denominator of a fraction, 
the subtraction of logarithms is conveniently avoided by the use 
of cologarithms. By our definition we have 

colog 25 = log 3jV = log 1 - log 25, Theorem III, § 209 
log 1 = 10. - 10 

log 25 = 1.3979 
colog 25= 8.6021-10 

Thus in dividing a number by 25 we may subtract the logarithm 
of 25, or what amounts to the same thing, add the logarithm of 
^ij, which is by definition the cologarithm of 26. 

EuLE. The cologarithm of any number is found by subtract- 
ing its logarithm from 10 — 10. 

In the process of division subtracting the logarithm of a num- 
ber and adding its cologarithm are equivalent operations. 



EXERCISES 



Compute, using cologarithms. 
- 8 X 62.73 X .052 



66 X 8.793 
Solution : 



2- MlVlf- 
4. V38.462 - 15.382. 
g 5086 (.0008769)8 
9802 (.001984)* * 



</: 



' 'XVsox 



log 8= .9031 

log 62. 73= 1.7975 

log. 052= 8.7160-10 

colog 56= 8.2518-10 

colog 8.793 = 9.0559 - 10 
log result = 27.7243 -30 

result = .00^99 



3. V1572 - 872. 

Hint . 1572 - 872 = (157 + 87) (157 - 87) 
= (244) (70). 



6. V(27:5)2 - (3.483)2. 



693 X .04692 
03841 X (569.8)2 



3 X 421.6 X 



.046.(200.1)^ 






(68.96)^ X 86.61 
.09263 V^ 

1416 X (5.638)2 
(75)i 



LOGARITHMS 247 

215. Change of base. We have seen that the logarithm of a 
number for the base 10 may be found to four decimal places in 
our tables. It is occasionally necessary to find the logarithm of a 
number for a base different from 10. For the sake of generality, 
we assume that the logarithms of all numbers for a base b are 
computed. We seek a means of finding the logarithm of any 
number, as x, for the base c ; that is, we seek to express log^a? in 
terms of logarithms for the base h. 

Suppose logc X = z, that is, c^ = x. 

Take the logarithm of this equation for the base h, and we have 

logftC^ = Z log^C = logftOJ. 



Then 


^ log,a^ 

l0g6« 


If we let 


M = l0gf,Gj 


we have 


.. ^og,x 



(1) 

This number M does not depend on the particular number x, 
but only on the two bases. From (1) we see that we can find the 
logarithm of any number for the base c by dividing its logarithm 
for the base b by M. The number M is called the modulus of the 
new system with respect to the original one. 

EuLE. To find the logarithm of a member for a new base c, 
divide the common logarithm by the modulus of the system whose 
base is c. 

EXERCISES 



Find: 




1. logs 21. 

solution: log321.>^-^; = lf^^f = 2.771. 
logioo .4771 




2. log5 6. 3. logalS. 


4. l0gi6 2. 


5. logs 167. 6. logi8 237. 


7. log2.i6l.41. 



248 



ADVANCED ALGEBRA 



N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


100 


0000 


0004 


0009 


0013 


0017 


0022 


0026 


0030 


0035 


0039 


101 
102 
103 

104 

105 
106 

107 
108 
109 


0043 
0086 
0128 

0170 
0212 
0253 

0294 
0334 
0374 


0048 
0090 
0133 

0175 
0216 
0257 

0298 
0338 
0378 


0052 
0095 
0137 

0179 
0220 
0261 

0302 
0342 
0382 


0056 
0099 
0141 

0183 
0224 
0265 

0306 
0346 
0386 


0060 
0103 
0145 

0187 
0228 
0269 

0310 
0350 
0390 


0065 

0107. 

0149 

0191 
0233 
0273 

0314 
0354 
0394 


0069 
0111 
0154 

0195 
0237 
0278 

0318 
0358 
0398 


0073 
0116 
0158 

0199 
0241 
0282 

0322 
0362 
0402 


0077 
0120 
0162 

0204 
0245 
0286 

0326 
0366 
0406 


0082 
0124 
0166 

0208 
0249 
0290 

0330 
0370 
0410 


110 


0414 


0418 


0422 


0426 


0430 


0434 


0438 


0441 


0445 


0449 


111 
112 
113 

114 
115 
116 

117 

118 
119 


0453 
0492 
0531 

0569 
0607 
0645 

0682 
0719 
0755 


0457 
0496 
0535 

0573 
0611 
0648 

0686 
0722 
0759 


0461 
0500 
0538 

0577 
0615 
0652 

0689 
0726 
0763 


0465 
0504 
0542 

0580 
0618 
0656 

0693 
0730 
0766 


0469 
0508 
0546 

0584 
0622 
0660 

0697 
0734 
0770 


0473 
0512 
0550 

0588 
0626 
0663 

0700 
0737 
0774 


0477 
0515 
0554 

0592 
0630 
0667 

0704 
0741 
0777 


0481 
0519 
0558 

0596 
0633 
0671 

0708 
0745 
0781 


0484 
0523 
0561 

0599 
0637 
0674 

0711 
0748 

0785 


0488 
0527 
0565 

0603 
0641 
0678 

0715 
0752 
0788 


120 


0792 


0795 


0799 


0803 


0806 


0810 


0813 


0817 


0821 


0824 


121 
122 
123 

124 

125 
126 

127 

128 
129 


0828 
0864 
0899 

0934 
0969 
1004 

1038 
1072 
1106 


0831 
0867 
0903 

0938 
0973 
1007 

1041 
1075 
1109 


0835 
0871 
0906 

0941 
0976 
1011 

1045 
1079 
1113 


0839 
0874 
0910 

0945 
0980 
1014 

1048 
1082 
1116 


0842 
0878 
0913 

0948 
0983 
1017 

1052 
1086 
1119 


0846 
0881 
0917 

0952 
0986 
1021 

1055 
1089 
1123 


0849 
0885 
0920 

0955 
0990 
1024 

1059 
1093 
1126 


0853 
0888 
0924 

0959 
0993 
1028 

1062 
1096 
1129 


0856 
0892 
0927 

0962 
0997 
1031 

1065 
1099 
1133 


0860 
0896 
0931 

0966 
1000 
1035 

1069 
1103 
1136 


130 


1139 


1143 


1146 


1149 


1153 


1156 


1159 


1163 


1166 


1169 


131 
132 
133 

134 
135 
136 

137 

138 
139 


1173 
1206 
1239 

1271 
1303 
1335 

1367 
1399 
1430 


1176 
1209 
1242 

1274 
1307 
1339 

1370 
1402 
1433 


1179 
1212 
1245 

1278 
1310 
1342 

1374 
1405 
1436 


1183 
1216 
1248 

1281 
1313 
1345 

1377 
1408 
1440 


1186 
1219 
1252 

1284 
1316 
1348 

1380 
1411 

1443 


1189 
1222 
1255 

1287 
1319 
1351 

1383 
1414 
1446 


1193 
1225 
1258 

1290 
1323 
1355 

1386 
1418 
1449 


1196 
1229 
1261 

1294 
1326 
1358 

1389 
1421 
1452 


1199 
1232 
1265 

1297 
1329 
1361 

1392 
1424 
1455 


1202 
1235 
1268 

1300 
1332 
1364 

1396 
1427 
1458 


140 


1461 


1464 


1467 


1471 


1474 


1477 


1480 


1483 


1486 


1489 


141 
142 
143 

144 
145 
146 

147 
148 
149 


1492 
1523 
1553 

1584 
1614 
1644 

1673 
1703 
1732 


1495 
1526 
1556 

1587 
1617 
1647 

1676 
1706 
1735 


1498 
1529 
1559 

1590 
1620 
1649 

. 1679 
1708 
1738 


1501 
1532 
1562 

1593 
1623 
1652 

1682 
1711 
1741 


1504 
1535 
1565 

1596 
1626 
1655 

1685 
1714 
1744 


1508 
1538 
1569 

1599 
1629 
1658 

1688 
1717 
1746 


1511 
1541 
1572 

1602 
1632 
1661 

1691 
1720 
1749 


1514 
1544 
1575 

1605 
1635 
1664 

1694 
1723 
1752 


1517 
1547 

1578 

1608 
1638 
1667 

1697 
1726 
1755 


1520 
1550 
1581 

1611 
1641 
1670 

1700 
1729 
1758 


150 


1761 


1764 


1767 


1770 


1772 


1775 


1778 


1781 


1784 


1787 


N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 



LOGARITHMS 



249 



N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


150 


1761 


1764 


1767 


1770 

1798 
1827 
1855 

1884 
1912 
1940 

1967 
1995 
2022 


1772 


1775 


1778 


1781 


1784 


1787 


151 
152 
153 

154 
155 
156 

157 

158 
159 


1790 
1818 
1847 

1875 
1903 
1931 

1959 
1987 
2014 


1793 
1821 
1850 

1878 
1906 
1934 

1962 
1989 
2017 


1796 
1824 
1853 

1881 
1909 
1937 

1965 
1992 
2019 


1801 
1830 
1858 

1886 
1915 
1942 

1970 
1998 
2025 


1804 
1833 
1861 

1889 
1917 
1945 

1973 
2000 
2028 


1807 
1836 
1864 

1892 
1920 
1948 

1976 
2003 
2030 


1810 
1838 
1867 

1895 
1923 
1951 

1978 
2006 
2033 


1813 
1841 
1870 

1898 
1926 
1953 

1981 
2009 
2036 


1816 
1844 
1872 

1901 
1928 
1956 

1984 
2011 
2038 


160 


2041 


2044 


2047 
2074 
2101 
2127 

2154 
2180 
2206 

2232 

2258 
2284 


2049 


2052 


2055 


2057 


2060 


2063 


2066 


161 
162 
163 

164 
165 
166 

167 
168 
160 


2068 
2095 
2122 

2148 
2175 
2201 

2227 
2253 
2279 


2071 
2098 
2125 

2151 
2177 
2204 

2230 
2256 
2281 


2076 
2103 
2130 

2156 
2183 
2209 

2235 
2261 

2287 


2079 
2106 
2133 

2159 
2185 
2212 

2238 
2263 
2289 


2082 
2109 
2135 

2162 
2188 
2214 

2240 
2266 
2292 


2084 
2111 
2138 

2164 
2191 
2217 

2243 
2269 
2294 


2087 
2114 
2140 

2167 
2193 
2219 

2245 
2271 

2297 


2090 
2117 
2143 

2170 
2196 
2222 

2248 
2274 
2299 


2092 
2119 
2146 

2172 
2198 
2225 

2251 
2276 
2302 


170 


2304 


2307 


2310 


2312 


2315 


2317 
2343 
2368 
2393 

2418 
2443 
2467 

2492 
2516 
2541 


2320 


2322 


2325 


2327 


171 
172 
173 

174 
175 
176 

177 

178 
179 


2330 
2355 
2380 

2405 
2430 
2455 

2480 
2504 
2529 


2333 
2358 
2383 

2408 
2433 
2458 

2482 
2507 
2551 


2335 
2360 
2385 

2410 
2435 
2460 

2485 
2509 
2533 


2338 
2363 
2388 

2413 
2438 
2463 

2487 
2512 
2536 


2340 
2365 
2390 

2415 
2440 
2465 

2490 
2514 

2538 


2345 
2370 
2395 

2420 
2445 
2470 

2494 
2519 
2543 


2348 
2373 
2398 

2423 
2448 
2472 

2497 
2521 
2545 


2350 
2375 
2400 

2425 
2450 
2475 

2499 
2524 
2548 


2353 
2378 
2403 

2428 
2453 
2477 

2502 
2526 
2550 


180 


2553 


2555 


2558 


2560 


2562 


2565 


2567 


2570 


2572 


2574 


181 
182 
183 

184 

185 
186 

187 
188 
189 


2577 
2601 
2625 

2648 
2672 
2695 

2718 
2742 
2765 


2579 
2603 
2627 

2651 
2674 
2697 

2721 
2744 
2767 


2582 
2605 
2629 

2653 
2676 
2700 

2723 
2746 
2769 


2584 
2608 
2632 

2655 
2679 
2702 

2725 

2749 

2772 


2586 
2610 
2634 

2658 
2681 
2704 

2728 
2751 
2774 


2589 
2613 
2636 

2660 
2683 
2707 

2730 
2753 
2776 


2591 
2615 
2639 

2662 
2686 
2709 

2732 
2755 

2778 


2594 
2617 
2641 

2665 
2688 
2711 

2735 

2758 
2781 


2596 
2620 
2643 

2667 
2690 
2714 

2737 
2760 
2783 


2598 
2622 
2646 

2669 
2693 
2716 

2739 
2762 
2785 


190 


2788 


2790 


2792 


2794 
2817 
2840 
2862 

2885 
2907 
2929 

2951 
2973 
2995 


2797 


2799 


2801 


2804 


2806 


2808 


191 
192 
193 

194 
195 
196 

197 
198 
199 


2810 
2833 
2856 

2878 
2900 
2923 

294.: 
2967 
2989 


2813 
2835 
2858 

2880 
2903 
2925 

2947 
2969 
2991 


2815 
2838 
2860 

2882 
2905 
2927 

2949 
2971 
2993 


2819 
2842 
2865 

2887 
2909 
2931 

2953 
2975 
2997 


2822 
2844 
2867 

2889 
2911 
2934 

2956 
2978 
2999 


2824 
2847 
2869 

2891 
2914 
2936 

2958 
2980 
3002 


2826 
2849 
2871 

2894 
2916 
2938 

2960 
2982 
3004 


2828 
2851 
2874 

2896 
2918 
2940 

2962 
2984 
3006 


2831 
2853 
2876 

2898 
2920 
2942 

2964 
2986 
3008 


200 


3010 


3012 


3015 


3017 


3019 


3021 


3023 


3025 


3028 


3030 


N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 



250 



ADVANCED ALGEBRA 



N. 





1 


2 3 


4 


5 


6 


7 


8 


M 


20 


3010 


3032 


3054 


3075 


3096 


3118 


3139 


3160 


3181 


320r 


21 

22 
23 

24 
25 
26 

27 
28 
29 


3222 
3424 
3617 

3802 
3979 
4150 

4314 
4472 
4624 


3243 
3444 
3636 

3820 
3997 
4166 

4330 
4487 
4639 


3263 
3464 
3655 

3838 
4014 
4183 

4346 
4502 
4654 


3284 
3483 
3674 

3856 
4031 
4200 

4362 
4518 
4669 


3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 


3324 
3522 
3711 

3892 
40(35 
4232 

4393 
4548 
4698 


3345 
3541 
3729 

3909 
4082 
4249 

4409 
4564 
4713 


3365 
3560 
3747 

3927 
4099 
4265 

4425 
4579 

4728 


3385 
3579 
3766 

3945 
4116 
4281 

4440 
4594 
4742 


3404 
3598 
3784- 

3962 
4133 
4298 

4456 
4609 
4757 


30 


4771 


4786 


4800 


4814 


4829 


4843 


4857 


4871 


4886 


4900 


31 
32 
33 

34 
35 
36 

37 
38 
39 


4914 
5051 
5185 

5315 
5441 
5563 

5682 
5798 
5911 


4928 
5065 
5198 

5328 
5453 
5575 

5694 
5809 
5922 


4942 
5079 
5211 

5340 
5465 
5587 

5705 
5821 
5933 


4955 
5092 
5224 

5353 
5478 
5599 

6717 
5832 
5944 


4969 
5105 
5237 

5366 
5490 
5611 

5729 
5843 
5955 


4983 
5119 
5250 

5378 
5502 
5623 

5740 

6855 
5966 


4997 
5132 
6263 

5391 
5514 
5635 

5752 
5866 
5977 


6011 
6146 
5276 

5403 
5527 

6647 

6763 

5877 
6988 


5024 
5159 
5289 

5416 
6639 
6658 

5775 
5888 
5999 


5038 
5172 
5302 

5428 
5551 
5670 

6786 
6899 
6010 


40 


6021 


6031 


6042 


6053 


6064 


6075 


6085 


6096 


6107 


6117 


41 
42 
43 

44 
45 
46 

47 

48 
49 


6128 
6232 
6335 

6435 
6532 
6628 

6721 
6812 
6902 


6138 
6243 
6345 

6444 
6542 
6637 

6730 
6821 
6911 


6149 
6253 
6355 

6454 
6551 
6646 

6739 
6830 
6920 


6160 
6263 
6365 

6464 
6561 
6656 

6749 
6839 
6928 


6170 
6274 
6375 

6474 
6571 
6665 

6758 
6848 
6937 


6180 
6284 
6385 

6484 
6580 
6675 

6767 
6857 
6946 


6191 
6294 
6395 

6493 
6590 
6684 

6776 
6866 
6955 


6201 
6304 
6405 

6503 
6599 
6693 

6786 
6875 
6964 


6212 
6314 
6415 

6513 
6609 
6702 

6794 
6884 
6972 


6222 
6325 
6425 

6522 
6618 
6712 

6803 
6893 
6981 


50 


6990 


6998 


7007 


7016 


7024 


7033 


7042 


7050 


7059 


7067 


51 
52 
53 

54 
55 
56 

57 

58 
59 


7076 
7160 
7243 

7324 
7404 

7482 

7559 
7634 
7709 


7084 
7168 
7251 

7332 
7412 
7490 

7566 
7642 
7716 


7093 
7177 
7259 

7340 
7419 
7497 

7574 
7649 
7723 


7101 
7185 
7267 

7348 
7427 
7505 

7582 
7657 
7731 


7110 
7193 
7275 

7356 
7435 
7513 

7589 
7664 
7738 


7118 
7202 
7284 

7364 
7443 

7620 

7597 
7672 
7745 


7126 
7210 
7292 

7372 
7451 

7528 

7604 
7679 
7752 


7135 
7218 
7300 

7380 
7469 
7636 

7612 
7686 
7760 


7143 
7226 
7308 

7388 
7466 
7543 

7619 
7694 
7767 


7152 
7235 
7316 

7396 

7474 
7551 

7627 
7701 
7774 


60 


7782 


7789 


7796 


7803 


7810 


7818 


7826 


7832 


7839 


7846 


61 
62 
63 

64 
65 
66 

67 
68 
69 


7853 
7924 
7993 

8062 
8129 
8195 

8261 
8325 
8388 


7860 
7931 
8000 

8069 
8136 
8202 

8267 
8331 
8395 


7868 
7938 
8007 

8075 
8142 
8209 

8274 
8338 
8401 


7875 
7945 
8014 

8082 
8149 
8215 

8280 
8344 
8407 


7882 
7952 
8021 

8089 
8156 

8222 

8287 
8351 
8414 


7889 
7959 
8028 

8096 
8162 
8228 

8293 
8367 
8420 


7896 
7966 
8036 

8102 
8169 
8235 

8299 
8363 
8426 


7903 
7973 
8041 

8109 
8176 
8241 

8306 
8370 
8432 


7910 
7980 
8048 

8116 
8182 
8248 

8312 
8376 
8439 


7917 
7987 
8055 

8122 

8189 
8254 

8319 
8382 
8445 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 



LOGARITHMS 



251 



N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


70 


8451 


8457 


8463 


8470 


8476 


8482 


8488 


8494 


8500 


8506 


71 
72 
73 


8513 
8573 
8633 


8519 
8579 
8639 


8525 

8585 
8645 


8531 
8591 
8651 


8537 
8597 

8657 


8543 
8603 
8663 


8549 
8609 
8669 


8555 
8615 
8675 


8561 
8621 
8681 


8667 
8627 
8686 


74 

75 
76 


8692 
8751 
8808 


8698 
8756 
8814 


8704 
8762 
8820 


8710 
8768 
8825 


8716 

8774 
8831 


8722 
8779 
8837 


8727 
8785 
8842 


8733 
8791 
8848 


8739 
8797 
8854 


8745 
8802 
8859 


77 
78 
79 


8865 
8921 
8976 


8871 
8927 
8982 


8876 
8932 
8987 


8882 
8938 
8993 


8887 
8943 
8998 


8893 
8949 
9004 


8899 
8954 
9009 


8904 
8960 
9015 


8910 
8965 
9020 


8915 
8971 
9025 


80 


9031 


9036 


9042 


9047 


9053 


9058 


9063 


9069 


9074 


9079 


81 
82 
83 


9085 
9138 
9191 


9090 
9143 
9196 


9096 
9149 
9201 


9101 
9154 
9206 


9106 
9159 
9212 


9112 
9165 
9217 


9117 
9170 
9222 


9122 
9175 
9227 


9128 
9180 
9232 


9133 
9186 
9238 


84 
85 
86 


9243 
9294 
9345 


9248 
9299 
9350 


9253 
9304 
9355 


9258 
9309 
9360 


9263 
9315 
9365 


9269 
9320 
9370 


9274 
9325 
9375 


9279 
9330 
9380 


9284 
9335 
9385 


9289 
9340 
9390 


87 
88 
89 


9395 
9445 
9494 


9400 
9450 
9499 


9405 
9455 
9504 


9410 
9460 
9509 


9415 
9465 
9513 


9420 
9469 
9518 


9425 
9474 
9523 


9430 
9479 
9528 


9435 
9484 
9533 


9440 
9489 
9538 


90 


9542 


9547 


9552 


9557 


9562 


9566 


9571 


9576 


9581 


9586 


91 
92 
93 


9590 
9638 
9685 


9595 
9643 
9689 


9600 
9647 
9694 


9605 
9652 
9699 


9609 
9657 
9703 


9614 
9661 
9708 


9619 
9666 
9713 


9624 
9671 
9717 


9628 
9675 
9722 


9633 
9680 
9727 


94 
95 
96 


9731 
9777 
9823 


9736 

9782 
9827 


9741 
9786 
9832 


9745 
9791 
9836 


9750 
9795 
9841 


9754 
9800 
9845 


9759 
9805 
9850 


9763 
9809 
9854 


9768 
9814 
9859 


9773 
9818 
9863 


97 

98 
99 


9868 
9912 
9956 


9872 
9917 
9961 


9877 
9921 
9965 


9881 
9926 
9969 


9886 
9930 
9974 


9890 
9934 
9978 


9894 
9939 
9983 


9899 
9943 
9987 


9903 
9948 
9991 


9908 
9952 
9996 


100 


0000 


0004 


0009 


0013 


0017 


0022 


0026 


0030 


0035 


0039 


N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 



216. Exponential equations. Equations in which the variable 
occurs only in the exponents may often be solved by the use of 
tables of logarithms if one keeps in mind the fact that 
log oc^ = ic log a. 



EXERCISES 

Solve the following : 

1. 10^-1 = 4. 

Solution : Taking the logarithm of both sides of the equation, we have 
(x-1) log 10 = log 4, 
or since log 10 = 1, a; = log 4 + 1 = .G021 + 1 = 1.0021. 



252 ADVANCED ALGEBRA 

2. 4* -3^ = 8, 
2* . 81/ = 9. 

Solution : Taking the logarithms of the equations, we have 

a;log4 + ylogS = log 8, 
a: log 2 + y log 8 = log 9, 

or jc21og2 4-ylog3 = 31og2, (1) 

a;log2 + y31og2 = 21og3. 
Eliminate x. 

x21og2+ 2/log3 = 31og2 

a;21og2 + y61og2 = 41og3 

2/(log3 - 61og2) = 31og2 - 41og3 

_ 31og2-41og3 _ 3 X .3010-4 x .4771 
^~ Iog3-01og2 ~ .4771 - 6 X .3010 

.9030 - 1.9084 _ - 1.0054 _ 1.0054 
~ .4771 - 1.8060 ~ - 1.3289 ~ 1.3289* 

Perform this division by logarithms. 

log 1.0054 = 10.0023 - 10 
log 1.3289= .1235 

logy= 9.8788-10 
y= .7565. 
Substituting in (1), 

a; = 31og2 - .7565 log3 _ .9030 - .7565 x .4771 
2 log 2 ~ .6020 

Compute .7666 x .4771 by logarithms. 

log. 7565= 9.8788-10 
log. 4771= 9.6786-10 . 
log result = 19. 5574 - 20 
result = .3609. 

„ .9030 -.3609 .5421 

Hence x — = 

.6020 .6020 

log.5421 = 19.7341 -20 

log. 6020= 9.7796 - 10 

log;c= 9.9545- 10 

X = .9005. 

3. 6* = 2. 4. 4y = 3. 5. 7*+« = 5. 

6. 32x + i_5. 7.4^-1 = 6^+1. 8. 22a:+3_ 6^-1 = 0. 

a^.6y = m, -^ a^.6!' = w, 

c* . d?/ = 71. ' x-\- y — n. 



LOGARITHMS 253 

-- 2^. 2^ = 222, -2 a2^-3.a3y-2:=a8. 

' x-y = A. 8x + 2y = n. 

- 3 3* • 4y = 15562, .. ^^2^^ . ^/^^F^i = a\ 

4^ . 52/ = 128000. 4,^:,— ^ 3 



^l)3x + 5 - ^sjb^y + 1 = 610. 

217. Compound interest. If |1500 is at the yearly interest 
of 3%, the total interest for a year is $1500 • (0.03) = $45. The 
total sum invested at the end of a year would be $1545. 

Let, in general, P represent a sum of money in dollars. 

Let r represent a yearly rate of interest. 

Then P • r represents the yearly interest on P, and 
P-f p.r = P(r + 1) 
represents the total investment, principal and interest, at the 
end of a year. 

Similarly, P (r + 1) r is the second year's interest, and 

P{r + l)r + P(r + 1) = P(r2 + r + r + 1) = ^(^ + 1)^ 
is the total investment at the end of two years. 

In general, A = P (r + 1)" (1) 

is the total accumulation at the end of tz years. If we know r, P, 
and n, we can by (1) find A. If we take the logarithm of both 
sides of the equation, we have 

log A = log P -\- n log (r + 1), 

log^-logP ^ (2) 

log(r + l) ^ ^ 

Hence if we know A, P, and r, we can find n. 
If the interest is computed semiannually, we have as interest 

v 

at the end of a half-year P • ^ ' while the entire sum would be 

P (- 4- 1 ) . Reasoning as above, we find that if the interest is com- 
puted semiannually, the accumulation at the end of n years is 

^=pg + l)" (3) 

Similarly, n= l"g -^ ' l»g f . (4) 



logg + l) 



254 ADVANCED ALGEBRA 



3 



If the interest is computed k times a year, we liave at the end 
of n years 

A-P(^ + 1) , (5) 



Hogg + l) 



EXERCISES 

In such exercises as the following, four-place tables are not sufficiently 
exact to obtain perfect accuracy. In general, the longer the term of years 
and the more frequent the compounding of interest, the greater the inaccuracy. 

1. If $1600 is placed at 3|% interest computed semiannually for 13 years, 
to how much will it amount in that time ? 



Hence 



ion: By formula (3), 


A = 


4 


-) 


2n 




P = 1600, r = 


.03|, 


n = 


13 




36 A = 1600 {-^ 


\26 


1600 


\400 


-)^=-Q^- 


log 1600= 3.2041 

26 log407 = 67.8496 

71.0537 

261og400 = 67.6546 

log^= 3.3991 

A = $2507. 










Iog407 = 2.e096 

26 

156576 

52192 

67.8496 

log 400 =2.6021 

26 

156126 

52042 



67.6546 

2. After how long will $600 at 6% computed annually amount to $1000 ? 

Solution : By formula (2) we have 

_ log J. — logP 

log(r + l) 

A = 1000, P = 600, r = .06. 

log 1000 - log 600 3 - 2.7782 .2218 _ ^,. 

n = -^ = =^ = 8./0 years. 

log 1.06 .0253 .0253 ^ 

.76 year = .76 • 12 = 9.12 months. 

.12 month = .12 • 30 = 3.6 days. 

Thus n = 8 years 9 months 3.6 days. 



LOGARITHMS 255 

In the following exercises the interest is computed annually unless the 
contrary is stated. 

3. To what will $3750 amount in 20 years if left at 6% interest ? 

4. To what sum will $25,300 amount in 10 years if left at 5% interest 
computed semiannually? 

5. To what does $1000 amount in 10 years if left at 6% interest computed 
(1) annually, (2) semiannually, (3) quarterly ? 

6. A sum of money is left 22 years at 4% and amounts to $17,000. How 
much was originally put at interest ? 

7. What sum of money left at 4|% for 30 years amounts to $30,000 ? 

8. What sum of money left 10 years at 4i% amounts to the same sum as 
$8549 left 7 years at 5% ? 

9. If a man left a certain sum 11 years at 4%, it would amount to $97 less 
than if he had left the same sum 9 years at 5%. What was the sum ? 

10. Which yields more, a sum left 10 years at 4% or 4 years at 10%? 
What is the difference for $1000 ? 

11. Two sums of money, $25,795 in all, are left 20 years at 4f%. The 
difference in the sums to which they amount is $14,660. What were the sums ? 

12. At what per cent interest must $15,000 be left in order to amount to 
$60,000 in 32 years ? 

13. At what per cent must $3333 be left so that in 24 years it will 
amount to $10,000 ? 

14. Two sums of which the second is double the first but is left at 2% less 
interest amount in 36 1 years to equal sums. At what per cent interest was 
each left ? 

15. In how many years will a sum double if left at 5% interest ? 

16. In how many years will a sum double if left at 6% interest computed 
semiannually ? 

17. In how many years will a sum amount to ten times itself if left at 
4% interest ? 

18. In how many years will $17,000 left at 4^% interest amount to the 
same as $7000 left at 5|% for 20 years ? 

19. On July 1, 1850, the sum of $1000 was left at 4i% interest. When 
paid back it amounted to $2222. When did this occur ? 

20. Prove formulas (1), (3), and (5) by complete induction. 



CHAPTEE XXI 
CONTINUED FRACTIONS 

218. Definitions. A fraction in the form 

e-¥f 

where a, &,•••, ^, •• • are real numbers, is called a continued fraction. 
We shall consider only those continued fractions in which tlie 
numerators b, d, /, etc., are equal to unity and in which the 
letters represent integers, as for example : 

»! H . written a^-\ — — 

When the number of quotients ^2? ^s? ^4? • • • is finite the frac- 
tion is said to be terminating. When the fraction is not terminat- 
ing it is infinite. We shall see that the character of the numbers 
represented by terminating fractions differs widely from that of 
the numbers represented by infinite continued fractions. We shall 
find, in fact, that any root of a linear equation in one variable, 
i.e. any rational number, may be represented by a terminating 
continued fraction, and conversely; furthermore, that any real 
irrational root of a quadratic equation may be represented by the 
simplest type of infinite continued fractions, and conversely. 

219. Terminating continued fractions. If we have a terminat- 
ing continued fraction, where ^i, a2j • • • are integers, it is evident 
that by reducing to its simplest form we obtain a rational num- 
ber. The converse is also true, as we can prove in the following 

256 



CONTINUED FRACTIONS 257 

Theorem. Any rational number may he expressed as a ter- 
minating fraction. 

Let - represent a rational number. Divide a by J, and let a^ 

be the quotient and c (which must be less than h) the remainder. 
Then (§ 26) , 

-^a, + - = a, + j 



Divide h by c, letting ^g t)® the quotient and d (which must be 
less than c) the remainder. Then 

- = «! H 

b a^ -\- d 

c 

Continuing this process, the maximum limit of the remainders 
in the successive divisions becomes smaller as we go on, until 
finally the remainder is zero. Hence the fraction is terminating. 
It is noted that the successive quotients are the denominators in 
the continued fraction. 

EXERCISES 

1. Convert the following into continued fractions 

(a) ^VV- 

Solution: 247|77[0 

77J247|3 
231 
16J77[4 
64 

13J16[1 
13 

3J13[4 
12 
1J313 
3 

The continued fraction is 

ZL-1 1 1 1 1. 

247 ~ 3 + 4 + 1 + 4 + 3* 



258 ADVANCED ALGEBRA 

(b) if. (c) ^Vt. (d) m- 

(e) iM- (f) Iff. (g) /jVf- 

(h) HH- (i) Hf. (J) mh 

2. Express the following continued fractions as rational fractious. 

^''^2 + 3 ^"^l + r ^"^-a + l- 

(d)l I. (e)i 11. (f)l 1 1. 

^ ' a; + x ^ M+2 + 3 ^ ' 3 + 4 + 5 

,.11111 ,^111111 

(g) ----- . (h) ------ . 

^^2+4+2+4+2 ^^1+2+3+1+2+3 

220. Convergents. The value obtained by taking only the first 
n ~ 1 quotients in a continued fraction is called the nth. convergent 
of the fraction. 

Thus iu the fraction 

1+1111 

2+3+2+6 : 

1 is the first convergent, 

1 3 

1 + - = - is the second convergent, 

■a 2t 

l + -=l+- = — is the third convergent, etc. 

3 
When there is no whole number preceding the fractional part of the continued 
fraction the first convergent is zero. Thus in 

111 

2 + 3 + 5 
\ is called the second convergent. 

In the continued fraction 

,1111 

^2 + ^3 + ^4 + «5 H 

let -i, -1, -i, . .. represent the successive convergents expressed 

9.\ ?2 ^z 
as rational fractions. 

Then for the first convergent we have 

^ = — > or^i = ai, g'i = l. 
9.\ 



CONTINUED FRACTIONS 259 

For the second convergent we have 

, 1 a^a^ + 1 J92 , H , ^ 

«! H = = — > or ^2 = «^2<^l + 1 = «2i?i + 1, 

^2 <^2 2'2 

!Z2 = «2 = ^22'l- 

For the third convergent we have 

^2 + 1^ a3«^2 + 1 «^3«2 +1 2'3 

«3 

or i?3 = % (^2% + 1) 4- ^1 = ^3^2 +i?l, 

?Z3 = «3«2 + 1 = %'Z2 + (1\- 

This indicates that the form of the r^th convergent is 

9'n a«2'n-l + 2'»-2 ^ 

This is in fact the case, as we proceed to show by complete 
induction. 

We have already established form (1) for n = 2 and n = ^. 
We assume it for n = m, and will show that its validity for 
n = m -\-l follows. The (m + l)th convergent differs from the 

mth only in the fact that a^ -\ appears in the continued 

fraction in place of a^. In (1) replace n by m, and a„ by 

a^ H J and we have 



Pr 



1+1 __ \ *^m+l/ 

'■^^ Um + ) 2'm-l + 2'm-2 

\ «'m+l/ 

_ (Q^m-H<^>n + ^)Pm^l + Q^„. + 1 J^m - 2 
(am+;iam + l).?m-l + «m + l?m-2 

_ Q^m4-lKi>m-l+i>m~2)+i>»»-l 
«^». + l(am2'm-l + 9'm-2)+ 2'm-l 



which is form (1). 



260 ADVANCED ALGEBRA 

EXERCISES 

1. Express the following as continued fractions, and find the convergents. 

(a) !?. 

Solution : By the method already explained, we find that 

30 _1 1 1 1 1 1 

il~l + 2+l + 2 + l + 2' 
Here ai = 0, a2 = 1, as = 2, 04 = 1, as = 2, aQ = 1, a? = 2. 
The first convergent is evidently 0, the second is 1, and the third is 
1 2 



mi- ir ^. ^ - P4 CliPS +P2 1-2+1 3 

The fourth convergent is — = -^^ — — = = -. 

Qi a^qs + ga 1-3 + 1 4 

mu ^x^u . • P5 ttsP* + P3 2-3 + 2 8 

The fifth convergent is — = —^ — ^ = = — • 

?5 a^qi + ^3 2-4 + 3 11 

aePn + P4 1-8 + 3 11 



The sixth convergent is ^^ = 
The seventh convergent is 



^6 tteO's + 5'4 1-11 + 4 15 
Pi OtPg+Ps 2-11 + 8 30 



(b) 3¥2. (c) If. (d) t¥t- (e) tVt- (f) t?7- 



Qi a^q& + q^ 2-15 + 11 41 

(f) \'-'- 

(g)TVj. (b)TW^. (i)?\V U) mV 

2. Find the value of the following by finding the successive convergents. 

,,11111 ,^,11111 

(a) - - - - -• (b) - - - - -. 

^^2+1+2+1+2 ^'3+2+3+2+3 

(c^ 1 i i i i i (d^ 1 1 1 1 1 1 

^^^2 + 3+1 + 1 + 3 + 2* ^^3 + 3_^34.34.34.3' 

(e^ 1 1 ' 1 1 1 1 (f\ 1 1 1 1 1 1 

^^^6 + 3 + 1 + 1 + 3 + 6' ^^1 + 34.5+5 + 3+1' 

1 1 1 1 1 1 

^^^ (X - 1) + X + (X + 1) * ^ ^ X + X + x' 

221. Recurring continued fractions. We have seen that e very- 
terminating continued fraction represents a rational number, and 
conversely. We now discuss the character of the numbers repre- 
sented by the simplest infinite continued fractions. A recurring 
continued fraction is one in which from a certain point on, a 
group of denominators is repeated in the same order. 



CONTINUED FRACTIONS 261 

----- ^ 

^^ 3 + 2 + 3 + 2 + 3+2 + *"' 

111111 

1+2+3+1+2+3+'" 

are recurring continued fractions if the denominators are assumed to repeat 
indefinitely as indicated. 

That a repeating continued fraction actually represents a num- 
ber we shall establish in § 223. Unless this fact is proven, one 
runs the risk of dealing with symbols which have no meaning. 
If for certain continued fractions the successive convergents 
increase without limit, or take on erratic values that approach no 
limit, it is important to discover the fact. All the fractions that 
we discuss actually represent numbers, as we shall see. 

We shall consider only continued fractions in which every 
denominator has a positive sign. 

Theorem. Every recurring continued fraction is the root of a 
quadratic equation. 

T w • . 111111 

Let, tor instance, a; = - - - - - - •••. 

' ' a -\- b -\- c -\- a -\- b -\- G -[- 

Evidently the part of the fraction after the first denominator c 
may be represented by cc, and we have thus virtually the termi- 
nating fraction 



X 



a -\- b -\- c -\- X 



The second convergent is — 
The third convergent is 

ab -\-l q^ 
The fourth convergent, or a?, gives us 

^jP4^ CT4^3+-^2 ^ (c+-a;)5+-l 
q^ ci^qz 4- S'2 (c + x) (ab +- 1) 4- fit 
Simplifying, we get 

(ab + l)x^-{- lc(ab -{-l)-{- a - b'jx - be -1 = 0, 



262 ADVANCED ALGEBRA 

which is a quadratic equation whose root is a?, the value of the 
continued fraction. 

Since this equation has a negative number for its constant term 
it has one positive and one negative root. The continued fraction 
must represent the positive root, since we assume that the letters 
la, h, c represent positive integers. The quadratic equation whose 
root is a recurring continued fraction with positive denomina- 
tors will always have one positive and one negative root. The 
equation will be quadratic, however, whatever the signs of the 
denominators may be. 

The proof may be extended to the case where there are any 
number of recurring denominators or any number of denominators 
before the recurrence sets in. Since every real irrational root 
of a quadratic equation is a surd, our result is equivalent to the 
statement that every recurring continued fraction may be ex jft*essed 
as a surd. 

EXERCISES 

Of what quadratic equations are the following roots ? Express the con- 
tinued fraction as a surd. 

ill 1 1 

* 2 + 3 + 2 + 3 + ""' 

Solution : Let x = - - . 

2 + 3 + a: 

Then x = l 1 ^ + ^ 



2+- 6+2X+1 

3 + « 



or 2 0:2 + 6 X - 3 = 0. 

Solving this equation, we get 

- 3 + VT6 - 3 - Vl5 
X\ = or X2 = 

2 2 

Since x^ is negative, xi must be the surd that is represented by the con- 
tinued fraction. 

1111 _3 + Vl6 



Thus 
2. 



2+3+2+3+ 2 

1111 3IIII 

1 + 2 + 1 + 2 + "*' ■ 3 + 2 + 3 + 2 + 



6.2 + 1 1 1 1 . 

2+1+2+1+ 


••• 


m.r.u. = 2 + l^\^. 


••. 


then ^-^=l^\^-- 




and . = 2 + i^-;^(^. 


-2) 



CONTINUED FRACTIONS 263 

^111111 silil 

• 1 + 2 + 3 + 1+2 + 3 + "" •3+1 + 3 + 1 + *"' 

« o 1 1 1 

7. 3 + - - - .... 

^3+4+5+ 

8.1 + 1 1 1 1 

3+4+3+4+ 

9. 1 + i 1 1 i i 1 .... 

1+2+3+1+2+3+ 

10. i 1 1 -1 1 i .... 

2+1+2+2+1+2+ 

222. Expression of a surd as a recurring continued fraction. 

This is the converse of the problem discussed in the last sec- 
tion, and shows that recurring continued fractions and quadratic 
equations are related in the same intimate way that terminat- 
ing fractions and rational numbers (i.e. the roots of linear equa- 
tions) are connected. We seek to express an irrational number, 
as, for instance, V2, as a continued fraction. This we may do 
as follows. 

Since 1 is the largest integer in V2 we may write 

V2 = l + (V2-l) = l+(-^^^^. 
Rationalizing the numerator, we have 

V2 = i + --i — 

V2+-1 

Since 2 is the largest integer in V2 +- 1 we have 

V2 = l+ 7^= 7 = 1 + 



2+-(V2-l) o I (V2-I) 

Rationalizing the numerator V2 — 1, we have 

^"^viTI ^ + 2+(V2-i) 



264 ADVANCED ALGEBRA 

By continuing this process we continually get tlie denomi- 
nator 2. Thus 11 

This process consists of the successive application of two opera- 
tions, and affords the 

EuLE. Express the surd as the sum of two numbers the first 
of which is the largest integer that it contains. 

nationalize the numerator of the fraction whose numerator is 
the second of these numbers. Repeat these operations until a 
recurrence of denominators is observed. 

This process may be applied to any surd, and a continued frac- 
tion which is recurring will always be obtained. We shall con- 
tent ourselves with a statement of this fact without proof. 

If the surd is of the form a — V^, a continued fraction may be 
derived for + V^ and its sign changed. Since the real roots of 
any quadratic equation cc^ + 2 a^x + ag = are surds of the form 
a ± V^, where a and b are integers, it appears that the roots of 
any such equation may be expressed as recurring continued frac- 
tions. It can be shown that the real roots of the general quad- 
ratic equation a-o^^ + ajcc -f ctg = may also be so expressed. 

EXERCISES 

1. Express the following surds as recurrent continued fractions, 
(a) 2 + Vs. 
Solution : 

2+V3 = 3 + (V3-l) = 3+ (1) 

= 34^^ = 3+ 2' 



V3+ 1 V3+I 

= 3 + ^^^ — = 3 + J = 3 + 



V3 + I ^ V3+I _ . V3-I 

2 .2 "^2 

3 + 1^ 3-1 =3 + ^^^ = 3 + 1 1 

2(V3 + 1) V3 + I 2-|-(V3-l) 

1 



conti:nued fkactions 265 

But since Vs — 1 is the same number that we have in (1), this fraction 
repeats from this point on, and we have 

2+V3 = 8 + l 1 I 1 .... 
1+2+1+2+ 



(b) V5. 


(c) Vl7. 




(d) V65. 




(e) V47. 


(f) Vli. 


(g) V23. 




(h) V3i. 




(i) Vl9. 


(J) V62. 


(k) V79. 




(1) V98. 




(m) V88. 


(n) V22. 


(0) Vis. 




(p) V59. 




(q) VlOl. 


(r) 7 + VII. 


(s) 8- 


-V3 




(t) 


3-V23. 



2. Express as a continued fraction the roots of the following equations, 
y (a) a;2 - 7 X - 3 = 0. / (b) x^ + 2 a; - 6 = 0. 

(c) x2 + 3x - 8 = 0. (d) x2 - 4x - 4 = 0. 

223. Properties of conver gents. The law of formation of con- 
vergents given in § 220 is valid whether the continued fraction 
is terminating or infinite. We should expect that in the case of 
an infinite fraction the successive convergents would give us an 
increasingly close approximation to the value of the fraction. 
This is indeed the fact, as we shall see. 

Theorem. The difference between the nth and (n + l)st con- 
vergents is ^ 

Mn + l 

We prove this theorem by complete induction. 
Let the continued fraction be 



a, H — — 

«2 + ^3 + »4 + 

Then the first and second convergents are respectively 
Then £2-a=(^i^L±l)_„, = i. 

2'2 S'l »2 «2 




(1) 



266 ADVANCED ALGEBRA 

Since ^i = 1? 2'2 = ^2? 

we have ^zl±i _^ = izL^yHl for n = 1. 

We assume that the theorem holds for n — m^ that is, 

Prn±l _ ^ -y^m^m + i+gmPm + l ^ {zlJ}!^. 

We must prove that it holds for n = m -{• 1, 

Now since - ^"^^^ — - ^'" + ^ ^ Pm + l^m + 2 YPm + 29^m+l^ 
2'm + 2 S'm + l 2'm + l2'»i + 2 

our theorem reduces to proving that the numerator 

-- Pm+lQm + 2 ■^Pm + 2^m + l=(- l)""^"- (2) 

In the left-hand member of (2) set 

«m + 2'7m+l + ^m = 2'm + 2, (1)> § 220 

and «^m + 2i^m + l+i^m=I>m + 2- 

Then -- Pm+li^m+^^m + l H- 2'm)f (^m+2P„.+ l +i?».)?n. + l 
=^m+l^m-^Pm<Im+l = - (Pm'Ln+ I - Pm + l^m) 

Corollary I. The difference between the successive convergents 
of a continued fraction with positive denominators approaches 
zero as a limit. 

Since q^ = a^q„_-^ + q^-^, evidently q^ increases without limit 
when n is increased, since to obtain a^ we add together positive 
numbers neither one of which can vanish. 

Thus we can find a value of n large enough so that — j and 

1 . ^« . 

hence ? will be smaller than any assigned number, which 

is another way of stating that as n increases approaches 

zero as a limit. ^'•^'* "*" ^ 



CONTINUED FliACTIONS 267 

Corollary II. The even convergents decrease, while the odd 
convergents increase, as n increases. 

We must show that 

PTn + 2 Pm 

9.m + 2 5'm 

is negative or positive according as ni is even or odd. Adding 
and subtracting ? we have 



?/m + 1 

Pm + 2 Pm (Pjn±2 Pin±\ i , 

m+l 



2'm + 2 ^m Vlm + 2 <! 



/Pjn±l_ln\ 
X^m + l ^mj 



_ (-1)"»+^ (-1)"* + ^ 

By Corollary I, the denominator of the first fraction exceeds 
that of the second. Hence when m is odd the sum in the last 
member of the equation is positive, and when 7n is even the sum 
is negative. 

We now se* that any recurring fraction of the type considered 
in § 221 actually represents a number in the sense of § 74. We 
have seen that the successive odd convergents continually increase, 
while the even convergents continually decrease, until the differ- 
ence between a pair of them is very small. Such sequences of 
numbers we have seen (§73 ff.) define real numbers. 

224. Limit of error. We are now in a position to state a maxi- 
mum value for the error made in taking any convergent of a con- 
tinued fraction for the fraction itself. 

Theorem. The maximum limit of error in taking the nth 
convergent for the continued fraction is less than 

Since by the theorem of the last section the value of the frac- 
tion is between any pair of consecutive convergents, it must 
differ from either of these convergents by less than they differ 

from each other, that is, by less than 



268 ADVANCED ALGEBRA 

EXERCISES 
Find a convergent that differs by less than .001 from each of the following ; 

1. V6. 
Solution : 

V6 = 2 + ( V6 - 2) = 2 + ^_~^ = 2 + 



V6 + 2 V6'+2 

2 

= 2 + i 6-4 =2 + 5 1 =^ + lj_l 

Since the last surd repeats the one in the first equation we have 

V6 = 2 + i 1 1 1 .... 

2+4+2+4+ 



Since 



Pi _ 2 . J92 _ 5 . Pa _ 4 • 5 + 2 _ 22 . 
gi ~ 1 ' 52 ~ 2 ' ^3 4 . 2 + 1 ~ 9 ' 




P4 2-22 + 5 49. ps 4-49 + 22 
g4~2.9 + 2~20' 95~4-20 + 9~ 


218 
89 


1-^-1 <.001. 





54^5 20 - 89 1780 

we see by § 224 that |f satisfies the condition, of the problem. 

2. V7. 3. V46. , 4. Vs. 5. Vl9. 

6. V36. 7. V32. 8. V6l. 9. V65. 

10. S+V2.- 11. V99. 12. Vil. 13. Vl3. 

14. The number ir has the value 3. 14169. Find by the method of continued 
fractions a series of convergents the last of which differs from this value by 
less than .OCioi. 



CHAPTEE XXII 

INEQUALITIES 

225. General theorems. We say that a is greater than h when 
a — 6 is positive. If a — h is negative, then a is less than h. 
Thus any positive number or zero is greater than any negative 
number. As we distinguished between identities and equations 
of condition in § 53, so in this discussion we observe that some 
statements of inequality are true for any real value of the letters, 
while others hold for particular values only. The former class 
may be called unconditional inequalities, the latter conditional. 

Thus a2 > — 1 is true for any real value of a and is unconditional, while 
a; — 1 > 2 only when a; is greater than 3 and is consequently conditional. 

The two inequalities <* > 5, g'> d are said to have the same 
sense. Similarly, a <b, c <d have the same sense. The inequal- 
ities a>h, c <id have a different sense. 

Theorem I. Any positive number may he added to, subtracted 
from, or midtiplied by both numbers of an inequality without 
affecting the sense of the inequality. 

Let a> b, that is, let a — b = k, where k is a positive number. 
If m is a positive number, evidently 





a ± m — (b ± m) = k, 


or 


a ±m> b±m. 


Similarly, 


ma — mb = mk, 


or 


ma > mb. 



The other statements of the theorem are proved similarly. 

Corollary. Terms may be transposed from one side of an 
inequality to the other as in the case of equations. 

269 



270 ADVANCED ALGEBRA 

Let a> b -{- G. 

Subtract c from both sides of the inequality and we obtain by 
Theorem I 

a — c> h. 

Theorem IL If the signs of both sides of an inequality are 
changed, the sense of the inequality must be reversed, that is, the 
> sign must be changed to <, or conversely. 

Let a> b, that is, let a — b = k, where ^ is a positive number. 

Then -a-\-b=-k, 

or (— «)—(—&) = — 7c, 

that is, by definition, — a < — b. 

EXERCISES 

Prove that the following identities are tnie for all real positive values of 

the letters. 

1. a2 + 62 > 2 ah. 

Solution : (a — h)^ is always positive. 

Thus a2 - 2 a6 + 62 = a2 + 62 _ 2 a6 is positive. 

That is, a2 + 62 > 2 ah. 

2. 3(a3 + 63)>a26 + a62. 

3. a2 + 62 + c2 > a6 + ac + he. 

4. (6 + c) (c + a) (a + 6) > 8 ahc. 

5. (a + 6 + c) (pfi + 62 + c2) > 9a6c. 

6. 62c2 + c2a2 + a262 > a6c (a + 6 + c). 

7. 3(a8 4- 63 + c8) > (a + 6 + c) {ah + 6c + ca). 

8. V(x + xi)2 + (y + 2/i)2 < Vx2 + 2/2 + y/x^^ + yi^. 

9. If a2 + 62 = 1, x2 + y2 = 1^ prove that ax-\-hy< 1. 

10. (a + 6 - c)2 + (a + c - 6)2 + (6 + c - a)2 > a6 + 6c 4- ca. 

11. Show that the sum of any positive number (except 1) and its reciprocal 
is greater than 2. 

12. Prove that the arithmetical mean of two unequal positive numbers 
always exceeds their geometrical mean. 



INEQUALITIES 271 

226. Conditional linear inequalities. If we wish to find the 
values of x for which 

ax -\- b < c, (1) 

where a, h, and c are numbers and a is positive, we may find 
such values by carrying out a process similar to that of solving a 
linear equation in one variable. 

By the corollary, § 225, we have from (1) 

ax < c — b. 



By Theorem I, § 225, x <- 



227. Conditional quadratic inequalities. We have already 
shown in § 116 that the quadratic expression ax^ -\- bx -\- c is posi- 
tive or negative, when the equation 

ax^ -{- bx -{- G = (1) 

has imaginary or equal roots, according as a is positive or nega- 
tive. If the equation has distinct real roots, the expression is 
positive or negative for values between those roots according as 
a is negative or positive. This we may express in tabular form 
as follows, for all values of x excepting the roots of (1), for 
which of course the expression vanishes. 



a 


62 _ 4 ac 


dx^ -\- bx + c 


+ 


-orO 


Always + 


- 


-orO 


Always — 


+ 


+ 


- for X between roots, + for other values 


- 


+ 


+ for X between roots, - for other values 



This enables us to answer immediately questions like the 
following : 

Example. For what values of cc is — 2 ic^+a; > — 3 ? By the corollary, § 225, 
this is equivalent to the question, For what value ofxis —2x^-^x-\-S>0? 

Here 62 — 4 ac = 1 -}- 24 = 25 is positive. The roots of the equation 
— 2 a;2 4- a; + 3 = are ic = — 1, a; = f . Thus by our table this expression 
is positive for all values of x between — 1 and |. 



272 ADVANCED ALGEBRA 

EXERCISES 

For what values of x are the following inequalities valid ? 

1. 2x-3>0. 

3. _x-l>7. 

- 9x 4 ^ 
3 7 

7. .12x.+ .3<1.3. 

9. 3<5x-2. 

11. x2-8x + 22>6. 
13. 2x2-3x>5. 
15. 2x2 -4x< -2. 
17. -3x2 + 2x<2. 
19. 5x2-8x<l. 
21. 3x2>3x-3. 



2. 


4x-7>l. 


4. 


- 3 X + 8 < 3. 


6. 


3 ^4^5 


8. 


3-4x>2. 


10. 


.s<i|-. 


12. 


x2 + 3x-2>l. 


14. 


-3x2-4x>8. 


16. 


3x2-9x>-6. 


18. 


- x2 + 6 X > 9. 


20. 


x2 < X - 1. 


22. 


3x>2x2-4. 



CHAPTEE XXIII 

VARIATION 

228. General principles. The number x is said to vary directly 
as the number y when the ratio of ic to 2/ is constant. , This we 
symbolize by 

i» oc 2/, or ^ = A^, (1) 

where A; is a constant. 

Thus if a man walks at a uniform speed, the distance that he 
goes varies directly as the time. If the length of the altitude of 
a triangle is given, the area of the triangle varies directly as the 
base. The volume of a sphere varies directly as the cube of 
its radius. 

The number x is said to vary inversely as the number y when 
X varies directly as the reciprocal of y. Thus x varies inversely 
as y when 

05 oc -5 OY - — xy =^ Uy (2) 

y i 
y 

where A: is a constant. Thus the speed of a horse might vary 
inversely as the weight of his load. The length of time to do 
a piece of work might vary inversely as the nmnber of laborers 
employed. 

The intensity of a light varies inversely as the square of the 
distance of the light from the point of observation. If I repre- 
sents the intensity of light and d the distance of the light from 
the point of observation, we have 

Zoc|. or| = ZcZ^ = A5, (3) 

where A; is a constant. - ' - 

273 



274 



ADVANCED ALGEBRA 



The number x is said to vary jointly as y and z when it varies 
directly as the product of y and z. Thus x varies jointly as y 
and z when 



X QC yz, or 



yz 



h 



(4) 



where Aj is a constant. 

Thus a man's wages might vary jointly as the number of days 
and the number of hours per day that he worked. 

The number x is said to vary directly as y and inversely as z 

when it varies directly with -• Thus the force of the attraction 

of gravitation between two bodies varies directly as their masses 
and inversely as the squares of their distances. If m represents 
the masses of two bodies, d their distance, and G the force of 
their attraction due to gravity, then 



m 
Goc^, or 



= k. 



(P) 



EXERCISES 

1. If a varies inversely as the square of 6, and if a = 2 when 6 = 3, what 
is the value of a when 6 is 18 ? 

Solution: By (3), a62 = k. 

We can determine k by substituting a = 2, 6 = 3. 

2 . 9 = A;. 
18 = A:. 



Then 



a . (18)2 = 18, 



or a = tV- 

2. The volume of a sphere varies as the cube of its radius. A sphere of 
radius 1 has a volume 4.19. What is the volume of a sphere of radius 3 ? 

Solution: Let F represent the volume and r the radius of the sphere. 
Then by (1), 

\ = k. 
4.19 



Determine k by substituting. 



Then 



= k. 



A; = 4. 19. 

V V 

-- = — = 4.19. 
(3)8 27 

F= 113.13. 



VARIATION 275 

3. ltxyxx + y, and a; = 1 when y = 1, find x when y = 8. 

4. The area of a circle varies as the square of the radius. If a circle of 
radius 1 has an area 3.14, find the area of a circle whose radius is 21. 

5. Find the volume of a sphere whose radius is .2. 
Hint. See exercise 2. 

6. The volume of a circular cylinder varies jointly with the altitude and 
the square of the radius of the base. A cylinder whose altitude and radius 
are each 1 has a volume of 3. 14. Find the volume of a cylinder whose 
altitude is 16 and whose radius is 3. 

7. The weight of a body of a given material varies directly with its 
volume. If a sphere of radius 1 inch weighs f of a pound, how much would 
a ball of the same material weigh whose radius is 16 inches ? 

8. The distance fallen by an object starting from rest varies as the square 
of the time of falling. If a body falls 16 feet in 1 second, how far will it 
fall in 6 seconds ? 

9. A body falls from the top to the bottom of a cliff in 3| seconds. How 
high is the cliff ? 

10. A triangle varies in area jointly as its base and altitude. The area 
of a triangle whose base and altitude are each 1 is ^. What is the area of a 
triangle whose base is 16 and altitude 7? 

11. If 6 men do a piece of work in 10 days, how long will it take 5 men 
to do it? 

12. If 3 men working 8 hours a day can finish a piece of work in 10 days, 
how many days will 8 men require if they work 9 hours a day ? 

13. An object is 30 feet from a light. To what point must it be moved in 
order to receive (a) half as much light, (b) three times as much light ? 

14. The weights of objects near the earth vary inversely as the squares 
of their distances from the center of the earth. The radius of the earth is 
about 4000 miles. If an object weighs 150 pounds on the surface of the 
earth, how much would it weigh 6000 miles distant from the center ? 



CHAPTEE XXIV 
PROBABILITY 

229. Illustration. If a bag contains 3 wlaite balls and 4 blact 
balls, and 1 ball is taken out at random, what is the chance that 
the ball drawn will be white ? 

This question we may answer as follows : There are 7 balls in 
the bag and we are as likely to get one as another. Thus a ball 
may be drawn in 7 different ways. Of these 7 possible ways 3 
will produce a white ball. Thus the chance that the ball drawn 
will be white is 3 to 7, or f. The chance that a black ball will 
be drawn is f . 

230. General statement. It is plain that we may generalize 
this illustration as follows : If an event may happen in ^ ways 
and fail in q ways, each way being equally probable, the chance 
or probability that it will happen in one of the ^ ways is 



V 



The chance that it will fail is 



p + q 



(1) 



(2) 



The sum of the chances of the event's happening and failing 
is 1, as we observe by adding (1) and (2). 

The odds in favor of the event are the ratio of the chance of 
happening to the chance of failure. In this case the odds in 
favor are 

I. (3) 

q 

The odds against the event are -. 

P 
276 



PKOBABILITi 277 



EXERCISES 

1. K the chance of an event's happening is ^^^ what are the odds in its 
favor ? 

v 1 
Solution : By (1), -^— = — . 

Hence lOjp = p + g, 

or 9p = g, 

«i 1 
or - = - , which by (3) are the odds in favor. 

q 9 

2. From a pack of 52 cards 3 are missing. What is the chance that they 
are all of a particular suit P 

Solution : The number of combinations of 62 cards taken 3 at a time is 

C52, 3 = — ' This represents p ■{■ q. The number of combinations of the 

1 * 2 • 3 13 • 12 • 11 

13 cards of any one suit taken 3 at a time is C13, 3 = ^ „ • This repre- 
sentep, 

13 • 12 • 11 

^ p 1-2.3 13.12-11 11 11 

Thus — =- — = = = ■ = . 

p-\-q 62 » 61 • 60 62.61-60 17-60 860 

1-2-3 

3. What is the chance of throwing one and only one 6 in a single throw 
of two dice ? 

Solution : There are 36 possible ways for the two dice to fall. This repre- 
sents p + g. Since a throw of two sixes is excluded there are 6 throws in 
which each die would be a 6, that is, 10 in all in which a 6 appears. This 
represents p. 

P 10 6 
Thus — - — = — = — 

p + q 36 18 

4. A bag contains 8 white and 12 black balls. What is the chance that a 
ball drawn shall be (a) white, (b) black ? 

5. A bag contains 4 red, 8 black, and 12 white balls. What is the chance 
that a ball drawn shall be (a) red, (b) white, (c) not black ? 

6. In the previous problem, if 3 balls are drawn, what is the chance that 
(a) all are black, (b) 2 red and 1 white ? 

7. What is the chance of throwing neither a 3 nor a 4 in a single throw 
of one die ? 

8. What is the chance in drawing a card from a pack that it be (a) an 
ace, (b) a diamond, (c) a face card ? 



278 ADVANCED ALGEBRA 

9. Three cards are missing from a pack. What is the chance that they 
are (a) of one color, (b) face cards, (c) aces ? 

10. A coin is tossed twice. What is the chance that heads will fall 
once? 

11. The chance that an event will happen is f. What are the odds in its 
favor ? 

12. The odds against the occurrence of an event are |. What is the chance 
of its happening ? 

13.' What is the chance of throwing 10 with a single throw of two dice ? 

14. A squad of 10 men stand in line. What is the chance that A and B 
are next each other ? 

15. What is the chance that in a game of whist a player has 6 trumps ? 

16. What is the chance that in a game of whist a player holds 4 aces ? 



CHAPTER XXV 
SCALES OF NOTATION 

231. General statement. The ordinary numbers with which 
we are acquainted are expressed by means of powers of 10. Thus 
263 = 2 . 10^ + 6 . 10^ + 3. 
This is the common scale of notation, and 10 is called the radix 
of the scale. 

In a similar manner a number might be expressed in any scale 
with any radix other than 10. If we take 6 as the radix, we shall 
have as a 'number in this scale, for instance, 
543 = 5 • 62 + 4 • 6 4- 3. 
In this scale we need only and five digits to express every 
positive integer. 

In general, if r is the radix of a scale of notation, any positive 
integer N will be denoted in this scale as follows : 

N=ao7^ + a^r--^ + a^r^-^ + . . . + o,^. (1) 

Theorem. Any positive . integer may he expressed in a scale 
of notation of radix r. 

Suppose we have a positive integer N. Let ?•" be the highest 
power of r that is contained in N. Then 

N = aoT- + iV'i, 
where N^ is less than r". Suppose that on dividing iVj by r"-^ we 
obtain ^^ ^ ^^^.„_i ^ ^^^ 

where N^ is less than r"~\ 

Then N = a^r^ + air«-i + N^. 

Proceeding in this manner we obtain finally 

N = aor"" -h 0^1^""^ H h a„, 

where the a's are positive integers less than r, or perhaps zeros. 

279 



280 ADVANCED ALGEBRA 

One observes that the symbol 10 indicates the radix in any 
system. In this general scale we need a and r — 1 digits to 
express every possible number. 

232. Fundamental operations. In the four fundamental operar 
tions in the common scale we carry and borrow 10 in computing. 
In computing in a scale of radix 6, for instance, we should carry 
and borrow 6. If the radix were r, we should carry or borrow r. 

Thus let r = 6. Then 4 + 5=1-6+3=13. Similarly, 5-3=2. 6 + 3 = 23. 
This is precisely analogous to our computation in the common scale, where, for 
instance, we would have 9 + 8 = 1 • 10 + 7 = 17, or 6 • 7 = 4 - 10 + 2 = 42. 

EXERCISES 

Perform the following operations. 

1. 2361 + 4253 + 2140 ; r = 7. 

2361 
4253 
2140 - 



12114 



In this process, since 3 + 1=4 and is less than the radix, there is nothing to carry. 
The next column gives 6 + 5 + 4= 15 =2-7+1, hence we write down 1 and carry 2. The 
next column gives 3 + 2+1 + 2=8=1-7 + 1, hence we write down 1 and carry 1. Finally 
we get 2 + 4 + 2 + 1 = 9=1-7 + 2, hence we write 12. 

2. 4602-3714; r = 8. 

4602 
3714 



Since we cannot take 4 from 2 we horrow one from the next place. Since the radix 
is 8 this amounts to 8 units in the first place. We then subtract 4 from 8 + 2, which 
leaves 6. In borrowing 1 from that digit is really reduced to 7 and the preceding digit 
to 5 ; then subtracting 1 from 7 we get 6. Since we cannot take 7 from 5 we borrow 8 
again and take 7 from 5 + 8= 13, which leaves 6. Since 1 has been borrowed from the 4 
we see the subtraction is complete since 3-3=0. 

3. 4321 . 432 ; r = 6. 

4821 

432 

14142 

24013 

33334 

4143222 

In multiplying by 2 we have nothing to carry until we multiply 3 by 2. This gives 
6=1-5 + 1. Hence we put down 1 and carry 1 to the product of 2 and 4. The addition of 
the partial products is carried out as in exercise 1. 



SCALES OF NOTATION 281 

4. 32130 H- 43 ; r = 6. 

43 1 32130 1 430 
300 
213 
213 
00 

In making an estimate for the first figure in the quotient we divide 32 by 4, keeping 
in mind that for this purpose 32=3-64-2. Thus 4 is contained in 20 just 5 times, but 
since our entire divisor is 43 we take 4 as the first figure in the quotient. The multipli- 
cations are of course performed as in exercise 3, excepting that here 6 is the radix. 

5. 4361 + 2635 + 5542 ; r = 1. 6. 5344 - 3456 ; r = 7. 
7. 2340 . 4101 ; r=6. 8. 6435 • 35 ; r = 7. 

9. 2003455 - 403 ; r = 6. 10. 344032 - 321 ; r = 5. 

11. 534401 - 443524 ; r = 6. 12. 425 + 254 + 542 + 452 ; r=6. 

233. Change of scale. If we have a number in the scale of 
radix r, we may find the expression for that number in the com- 
mon scale by writing the number in form (1), § 231, and carrying 
out the indicated operations.. 

Example. Convert 4635, where r = 7, into the ordinary scale. 
4635 = 4- 73 + 6. 72 + 3-7+5 
= 4 • 343 + 6 . 40 + 3 . 7 + 5 
= 1692. 

If we have a number in the common scale, we may express it 
in the scale with radix r as follows : If the number is N, we have 
to determine the integers aQ, a^, • • , a„ in the expression 

N = ao7^ + «^i^""^ H h a^-iT + ««• (1) 

Divide (1) by r. We have 

:^= aor"-i + a,r-' H- a^,_, + - = iV' +-; 
r r r 

that is, the remainder a^ of this division is the last digit in the 
expression desired. 

Divide N' by r and we obtain 

— = JSJ-" = a^r^-^ + a^r^-^ H -|- ^^^^ ; 

r r 



282 ADVANCED ALGEBRA 

that is, the remainder from this division is the next to the last 
digit in the desired expression. Proceeding in this way we obtain 
all the digits fl^„, c^„ _ i, • • • , ^i, «o- 

Example. Express 37496 in the scale with radix 7. 

7) 37496 

7 ) 5356 remainder 4 

1 ) 765 remainder 1 

• 7 ) 109 remainder 2 

7 ) 15 remainder 4 

7 )2 remainder 1 

remainder 2 

The number in scale r = 7 is 214214. 

To change a number from any scale rj to any other scale r^j 
we may first change the number to the scale of 10 and then by 
the process just given to the scale r^. The process indicated in 
the preceding example may be employed directly to change from 
any scale to any other, provided the division is carried out in the 
scale in which the number is given. One of these methods may 
be used to check the other. 

Example. Change 34503 from scale r = 6 to one in which r = 9. 

34503 = 3. 6* + 4- 63 +5. 62 + 3 = 4935 in scale of 10. ♦ 

9 )4935 

9 ) 548 remainder 3 

9 )60 remainder 8 

9)6 remainder 6 

remainder 6 

Thus 34503 in scale of 6 becomes 6683 in scale of 9. 

Check: 9)34503 

9)2312 remainder 3 ^" carrying out this division it must 

■ X • A Q ^® kept in mind that the dividends are in 

92140 remamder 8 g^^jg ^f g^ ^l^jle ^^Q remainders are to be 

9) 10 remainder 6 in scale of 9. 
remainder 6 

234. Fractions. In the ordinary notation we express fractional 
numbers by digits following the decimal point. This notation 
may also be used in a scale with any radix. 



SCALES OF NOTATION 283 

Thus the expression .5421 stands for 

10 102 ^ 10« 10* 
in the common scale. 

In the scale with radix r it stands for 

ry% ry%** n%9 niT 

The process of changing the scale for fractions is performed 
in accordance with the same principles as are employed in the 
change of scale for integers. The following examples suffice to 
illustrate it. 

Example 1. Express .5421 in the scale of 6 as a decimal fraction. 

.6421 = 5 + 1+1+1 
6 62 68 64 

5.63 + 4.62+2-6+1 1237 ^,,, 
6* 1296 

Example 2. Express .439 as a fraction for radix 6. 

Let .439 = ^ + A + l + l + .... 

6 62 68 6* 

Multiplying by 6, 2.634 = a + ^ + ^ + ^ + • . . . 

b o-' 6' 

Thus a = 2 and we have .634 = - + — f- 1 . 

6 ^ 62 68 

c d 
Multiplying by 6, 3.804 = 6 + - + — + ... . 

6 62 

c d 
Thus 6 = 3 and we have .804 = - + — + ••. . 

6 62^ 

Multiplying by 6, 4.824 = c + - + •• .. 

6 

d 
Thus c = 4 and we have .824 = - + .... 

6 

Multiplying by (T, 4.944 = d + • . .. 

Thus d = 4. 

The fraction in scale of radix 6 is then .2344 • • «„ 



284 ADVANCED ALGEBRA 

EXERCISES 

1. Express the following as decimal fractions, 
(a) .374; r = 8. (b) .4352; r = 6. 
(c) .2231 ; r = 4. (d) .2001 ; r = 3. 

2. Express the decimal fraction .296 as a radix fraction for r = 5. 

3. Express the decimal fraction .3405 as a radix fraction for r = Q. 

34 

4. Express — as a radix fraction for r = 4. 

^ 128 

5. Express as a radix fraction for r = 5. 

^ 626 

6. In what scale is 42 expressed as 1120? 

Solution : We seek r where 

r3 + r2 + 2 r = 42. 

This is equivalent to finding a positive integral root of the equation 

r3 + r2 + 2 r - 42 = 0. 
By synthetic division, 

1 + 1 + 2 - 42(2 
+ 2 + 6 + 16 
+ 3 4- 8-26 
1 + 1+ 2-4213 
+ 3 + 12 + 42 
+ 4+14 
Thus 3 is the value sought. 

Check : 3^ + 32 + 2 • 3 = 27 + 9 + 6 = 42. 

7. In what scale is 2704 denoted by 20304 ? 

8. In what scale is 256 denoted by 10000 ? 

9. In what scale is .1664 denoted by .0404 ? 
10. Show that 1331 is a perfect cube. 

235. Duodecimals. We may apply some of the foregoing 
processes to mensuration. If we take one foot as a unit and the 
radix as 12, we may express distances in the so-called duodecimal 
notation. Thus 2 ft. 6 in. is represented in the duodecimal scale 
by 2.6. Since in a scale of radix r we need ?• — 1 symbols, we 
will let 10 = ^ and 11 = e. Thus 21 ft. 10 in. would be expressed 
in duodecimal notation as 19.^. We may now find areas and vol- 
umes in this notation much more readily than by the usual method 
of converting all distances to inches. 



SCALES OF NOTATION 285 

Example. Multiply 8 ft. 3 in. by 3 ft. 10 in. We multiply 8 • 3 by 3.i 

8.3 in the duodecimal scale. To convert the result to square 

3-t feet and square inches we must keep in mind that 27.76 

6t6 ^ 2 . 12 + 7 + tV + xf^ = 31 sq. ft. 90 sq. in., since 144 

— — square inches equal one square foot. 

This example suggests the following method of multiplying 
distances : 

EuLE. Express the distances in duodecimal notation with the 
foot as a unit. 

Multiply in the scale for which r = 12. 

In the product change the part on the left of the point from 
duodecimal to decimal scale. 

Multiply the digit following the point hy 12^ and add to the 
last figure to obtain the square inches in the result. 

EXERCISES 

1. Multiply the following : 

(a) 13 ft. 4 in. by 67 ft. 11 in. 

Solution: 11.4 

57.e 
1028 
794 

568 



636.68 = 905 sq. ft. 80 sq. in. 

(b) 10 ft. 6 in. by 12 ft. 2 in. 

(c) 8 ft. 4 in. by 11 ft. 11 in. 

(d) 23 ft. 6 in. by 47 ft. 8 in. 

(e) 41 ft. 6 in. by 36 ft. 1 in. 

2. What is the area of a room 16 ft. 2 in. by 10 ft. 3 in. ? 

3. What is the area of a walk 60 ft. 6 in. by 3 ft. 3 in. ? 

4. What is the area of a city lot 62 ft. 6 in. by 163 ft. 7 in. ? 






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